当前位置：文档库 › Supersimple fields and division rings

Supersimple fields and division rings

a r

X i

t h /

[

t h

]

e p

8Supersimple ?elds and division rings A.Pillay ?University of Illinois and MSRI T.Scanlon ?MSRI F.O.Wagner ?University of Oxford and MSRI February 1,2008Abstract It is proved that any supersimple ?eld has trivial Brauer group,and more generally that any supersimple division ring is commutative.As prerequisites we prove several results about generic types in groups and ?elds whose theory is simple.1Introduction Simple theories were introduced by Shelah in [12].In [3],Kim,continuing Shelah’s work,showed how the theory of forking transfers almost completely from the stable context to the more general simple context.In [4]the “In-dependence Theorem”was proved for simple theories,giving a satisfactory analogue of the theory of stationarity from stability theory.

In [8],Poizat’s theory ([10])of generic types and stabilizers in stable groups was generalized to the case of groups de?nable in simple theories.Further generalizations appear in [13].

Stability-theoretic algebra studied,among other things,the algebraic con-sequences of imposing stability-theoretic conditions on a group,ring or?eld. Among the main results in the area was the Macintyre-Cherlin-Shelah theo-rem([7],[1])saying that any in?nite?eld whose theory is superstable must be algebraically closed.

Supersimple?elds(namely?elds with supersimple theory)on the other hand form a broader class.In[5]it is shown that any perfect pseudo-algebraically closed(P AC)?eld with“small”absolute Galois group is su-persimple;in fact of SU-rank1in the language of rings.(Ultraproducts of ?nite?elds furnish examples.)The work in this paper is partly motivated by the conjecture that these are the only cases.It follows from[9]that any su-persimple?eld is perfect and has small absolute Galois group.In the present paper we show that a supersimple?eld K has trivial Brauer group.As su-persimplicity is preserved under?nite extensions,this is,by[11],X.Prop.11, equivalent to the norm map N:L?→K?being surjective for any?nite Galois extension L of K.Triviality of the Brauer group also implies that any rational variety over K has a K-rational point,yielding in a sense a?rst approximation to the conjecture that any supersimple?eld must be P AC.

We assume acquaintance with the notions and machinery from[4]and [8].In section2,some additional results are obtained about generic types and connected components of groups in simple theories.In section3,we show that if K is a?eld in a simple theory then the notions of additive and multiplicative generic coincide.We also make some observations about the interaction between the additive and multiplicative connected components. These results are applied in section4to show triviality of the Brauer group in the supersimple case.Some additional algebraic arguments yield in section 5commutativity of any supersimple division ring.

The?rst author would like to thank Zoe Chatzidakis for many fruitful discussions around the time when he?rst began considering the issues dealt with in this paper.In particular the result that if K is a supersimple?eld of characteristic0,then any rational curve over K has a K-rational point,was obtained together with her.

The result on the triviality of the Brauer group was obtained by all three authors,and the results in section5on division rings were obtained by Wag-ner.

2More on generic types and stabilizers

Let us?x a saturated modelˉM of a simple theory,and an in?nite group G which is type-de?nable inˉM over?say.All complete types p(x)we consider will be types of elements of G(namely we assume“p(x)→x∈G”).We will often wish to work with Lascar strong types(so that we can use the Independence Theorem).This is most easily accomplished by working with types over models.So M will denote a small elementary substructure ofˉM, which may vary.A,B as usual denote small subsets ofˉM.

We brie?y recall(from[8])the notions of generic type,stabilizer,and connected component,

Generics.Let a∈G.Then tp(a/A)is(left-)generic(for G)if whenever b∈G is independent from a over A then b.a is independent from A∪{b}over ?.(We also say that a is a generic element of G over A.)We have associated notions of a(type)-de?nable subset of G being generic in G.Moreover the notions of left-generic and right-generic coincide.

Stabilizers.For p(x)∈S(M),St(p)is de?ned to be{b∈G:for some realization a of p which is independent from b over M,b.a realises p(and is independent from b over M}.St(p)is type-de?nable over M.Let Stab(p) be the subgroup of G generated by St(p).Then it turns out that Stab(p)is type-de?nable over M,and moreover that St(p)is“large”in Stab(p)in the sense that any generic element of Stab(p)over M is in St(p).

Remark2.1One can also de?ne St(p,q)for p(x),q(x)∈S(M)in the obvi-ous way:St(p,q)={b∈G:for some a realising p such that a is independent from b over M,b.a realises q and is independent from b over M}.St(p,q) is again type-de?nable over M and one sees easily,using the Independence Theorem,that St(p,q)is,if nonempty a large subset of a translate of Stab(p).

Connected components.For any set A of parameters,G0

A ,the connected

component of G over A,is by de?nition the smallest type-de?nable over A subgroup of G of bounded index.A key fact is that p(x)∈S(M)is a generic type of G if and only if Stab(p)contains G0M.Also clearly any generic type of G0A is a generic type of G.In contrast to the stable situation,the connected component of G over A may vary with A.On the other hand G0A is always a normal subgroup of G.

If G happens to be stable then we know that G0has a unique generic type,whose stabilizer is G0.The next lemma gives an analogue of this in the simple case.

Proposition2.2Let p(x),q(x),r(x)∈S(M)be generic types of G0

M .Then

there are realizations a,b,c of p,q,r respectively,which are pairwise M-independent,and with a.b=c.

Proof.This can be seen using Remark2.1,but we give a direct proof.Let b,c be independent(over M)realizations of q,r respectively.Let a′=c.b?1. Then a′.b=c,a′is generic in G0M over M and{a′,b,c}is pairwise M-independent.We can similarly?nd d generic in G0M over M,such that a=a′.d realises p and{d,a′,a}is pairwise M-independent.By the facts above on connected components and generics,Stab(q)contains G0M and hence any generic element of G0M over M is in St(q).In particular d?1∈St(q). By the Independence Theorem we can assume that{d,a′,b}is pairwise M-independent and that d?1.b=b′realises q.Then(a′.d).b′=c′realises r. That is a.b′=c,and easily{a,b′,c}is pairwise M-independent.

3Simple?elds

In this section F will be an in?nite?eld which is type-de?nable over?in the saturated modelˉM of a simple theory T.Types p(x),q(x)will be types of elements of F.We are going to apply Proposition2.2in order to understand the interaction between generic additive subgroups and generic multiplicative subgroups in simple?elds.(In the stable case the situation is clear:a stable ?eld is connected both additively and multiplicatively.)We begin by pointing out that in simple?elds additive and multiplicative generics coincide. Proposition3.1Let F be an in?nite?eld which is type-de?nable inˉM over?.Then a type p(x)∈S(A)is an additive generic type of F i?it is a multiplicative generic type of F.

Proof.We begin with

Claim1.If p(x)∈S(A)is additively generic then for any nonzero a∈F and realisation c of p independent from a over A,tp(a.c/A∪{a})is an additive generic of F.

Proof of Claim1.Let a∈F and M contain A∪{a}.Let p′be a nonforking extension of p over M,realised by b.Then p′remains an additive generic of F.So the additive stabilizer of p′,Stab+(p′)is a type-de?nable over M subgroup of F+of bounded index.It follows that Stab+(a.p′)=a.Stab+(p′) also has bounded index.By the remarks above on connected components a.p′is an additive generic type of F.In particular a.p′does not fork over?. From Claim1,using the fact that tp(a.c/A∪{a})there,being an additive generic,does not fork over?,we see that:

Claim2.Any additive generic type is a multiplicative generic type.

Claim3.Any multiplicative generic type of F is an additive generic type of F.

Proof of Claim3.Let q(x)∈S(A)be a multiplicative generic type.We may replace q any time by a nonforking extension.By Claim2(and the existence of generic types)there is an additive generic type p(x)∈S(A)which is also multiplicatively generic.It follows that there is c∈F,M containing A∪{c} and nonforking extensions q′,p′of q,p over M such that c.q′=p′.So c.q′is an additive generic type.By Claim1,q′is also an additive generic type. Lemma3.2Let T be a multiplicative subgroup of F of bounded index,type-de?nable over A.Then every nonzero coset a.T of T meets(F+)0A in a generic set.

Proof.Note that if B?A,then(F+)0B is contained in(F+)0A.So we may assume that A is a model M which contains a representative of each coset c.T of T in F?.Note that(*)For each nonzero c∈F M,c.(F+)0M=(F+)0M (as multiplication by c is an additive endomorphism of F which interchanges M-de?nable sets).Let p(x)be a generic type of G0M over M.Let S be a coset of T in F?.Then there is c∈F M such that c.p is the type of an element of S.By Claim1in the proof of Proposition3.1,together with(*),c.p is a generic type of(F+)0M.This proves the lemma.

Remark3.3In fact,with the assumptions of Lemma3.2,every coset of (F+)0A in F+meets every coset of T in F?in a generic set.

Proof.We may assume that A is a model M which contains representatives of all cosets of T in F?.To prove the remark,it is enough(using Lemma 3.2)to show that if S is a coset of(F+)0M in F+then S∩T is generic.Let

b∈S be nonzero.Let M′be a model containing M∪{b}.By Lemma3.2, we can?nd generic d in(F+)0M′over M′such that d∈b?1.T.Let c=d?1. Then c is generic in(F+)0M′over M′too,and as in the proof of Lemma3.2, b.c is generic in(F+)0M′over M′.So b.d=b.c+b is generic in S over M.On

the other hand b.d∈T.Thus S∩T is generic,as required.

Proposition3.4Let T be any multiplicative subgroup of F of bounded in-dex.Let S1,S2be cosets of T in F?.Then S1+S2(the set of s1+s2for s1∈S1and s2∈S2)contains F?.In fact for any nonzero d∈F there are a,b∈S1,S2respectively,each generic over d such that a+b=d.

Proof.Again we work over a reasonably saturated model M(so in particular T,S1,S2are de?ned over M).Let d∈F M be nonzero,and let S3be d.T. By Lemma3.2,let p(x),q(x),r(x)be generic types of(F+)0M such that p(x)→x∈S1,q(x)→x∈S2and r(x)→x∈S3.By Proposition2.2, there are realisations a,b,c of p,q,r respectively,such that a+b=c,where moreover{a,b,c}is pairwise independent over M.Multiplying on the right by c?1.d∈T,we obtain a′∈S1,b′∈S2with a′+b′=d.Clearly each of a′,b′are generic over M.This completes the proof.

[Here is a sketch of another proof of3.4,avoiding use of Proposition2.2 and containing some possibly useful ideas:Work again over a model M. Let X=St(S1,S2)which by de?nition equals{a∈F:for each generic type p(x)∈S(M)of S1there is c realising p independent from a such that a.c∈S2}.X is type-de?nable over M and moreover(see2.2)there is a type-de?nable set X′,a union of cosets of(F+)0M in F+such that X?X′and also every generic element of X′over M is in X.Note that X is invariant under multiplication by T.Let now H be the set-theoretic stabilizer of X′.

H is a type-de?nable over M subgroup of F+containing(F+)0M.H is also invariant under multiplication by T,but we know that(F+)0M intersects each coset of T in F?.Thus H=F+.It follows that X′=F,and thus X contains all generic types over M.As X is also invariant under T it follows that X contains F?.]

The characteristic0case of the following was observed earlier together with Zoe Chatzidakis.

Corollary3.5If F is supersimple,then any conic de?ned over F has an F-rational point.

Proof.A conic de?ned over F can be put into the form x2+ay2=b for some nonzero a,b∈F.As we note in the next section,supersimplicity of F implies that the squares form a de?nable subgroup of F?of?nite index.By 3.4,the equation has a solution in F.

4Triviality of the Brauer group for super-simple?elds

We will prove that if F is an in?nite?eld possibly with extra structure,whose theory is supersimple,then F has trivial Brauer group.We will see in the course of the proof that this is a?rst order property,so we will assume F to be saturated.We will work with the formalism of the previous section, namely we assume F to be de?nable over?in a big modelˉM of a supersimple theory.

At this point it is worth mention the relevant facts about the SU-rank, most of which will be used in the next section.Each complete type has ordinal valued SU-rank,and the generic types of a(type-)de?nable group G are precisely the types of maximal SU-rank in G(which exist).The following is mentioned explicitly in[13]

Fact4.1(i)for any elements a,b and set A of parameters,SU(a/A∪{b})+ SU(b/A)≤SU(a,b/A)≤SU(a/A∪{b})⊕SU(b/A).

(ii)Let G be a type-de?nable group and H a type-de?nable subgroup,then SU(H)+SU(G/H)≤SU(G)≤SU(H)⊕SU(G/H)

(iii)Suppose D is a type-de?nable division ring,then D is de?nable and has monomial SU-rank.

The relevant consequences for this section are:

Fact4.2(i)for each n,the group of nth.powers(F?)n has?nite index in F?,

(ii)F is perfect.

Proof.(i)Let a∈F be generic over?.Let b=a n.Then a∈acl(b),so SU(a/b)=0.Thus by Fact4.1.(i)SU(a)=SU(b),whereby b is generic in F too.Thus(F?)n is a generic subgroup of F?and has?nite index(as it is de?nable).

(ii)Supposing F to have characteristic p>0,note that as in(i)the additive group of F p has?nite index in F.But F/F p is a vector space over the?eld F p,a contradiction(the latter being in?nite)unless F p=F.

Note that any?nite extension of F is also de?nable inˉM,so Fact4.2applies to all?nite extensions of F.

We will now recall relevant notions and facts concerning the Brauer group, the norm map and Galois cohomology.In fact one can extract a general result which says that for the Brauer group of every?nite extension of a?eld F to be trivial it is enough that for any?nite extension K of F and Kummer extension L of K,N L/K is surjective.(Here it is assumed that every?nite extension of F is perfect.)Proposition3.4will allow us to conclude triviality of the Brauer group for supersimple F.Rather than simply state this reduction, we will include an explanation of it as part of the proof,which will entail giving some de?nitions.

We?rst discuss the Brauer group.The reader can look at[2]and[11] for further details.We will assume that F and all its?nite extensions are perfect.

By a central simple algebra over F we mean a?nite dimensional F-algebra A whose centre is F and which has no nontrivial two-sided ideals.If A and B are two such objects then so is the tensor product A?F B.A and B are called equivalent(or similar)if for some m,n the matrix algebras M m(A),M n(B) are isomorphic(as F-algebras).The tensor product operation respects this equivalence relation and turns the set of classes into an abelian group.This group is called the Brauer group of F,Br(F),and is an important invariant of the?eld F.Any central simple F-algebra A will be a matrix algebra over a certain?nite dimensional division algebra D with centre F.Moreover the equivalence class of A is determined by and determines the isomorphism type of D.The trivial element of Br(F)then corresponds to F itself.On the other hand,for any central simple algebra A over F there i s some?nite extension K of F such that A?K is isomorphic to a matrix algebra over K, hence represents the trivial element of Br(K).(K is called a splitting?eld for A.)

If K is a?nite extension of F then tensoring with K determines a ho-momorphism from Br(F)into Br(K).The kernel of this homomorphism is denoted Br(K/F).From the previous paragraph,Br(F)is the union of all Br(K/F)as K runs over?nite extensions of F,in fact,over?nite Galois extensions of F.

For K a?nite Galois extension of F with Galois group G,there is a classi-cal isomorphism of Br(K/F)with the Galois cohomology group H2(G,K?): given a2-cocycle f:G×G→K?,de?ne a F-algebra structure on the K-vector space A with basis{u s:s∈G}by:u s·u t=f(s,t)u st and for e∈K, u s e=(se)u s.A then becomes a central simple F-algebra A f,and the map f→A f determines an isomorphism between H2(G,K?)and Br(K/F).We have the following fact(see[11],X.6):

Fact4.3Let F

0→H2(H,K?)→H2(G,L?)→H2(Gal(L/K),L?).

We also need([11],IX):

Fact4.4Let G be a?nite group and A a G-module.Let n≥1.Suppose that for all primes p,H n(G p,A)=0where G p is a Sylow subgroup of G. Then H n(G,A)=0.

Finally,for K a?nite Galois extension of F,the norm map N K/F:K?→F?is the map which takes any a∈K?to the product of all sa where s runs over Gal(K/F).See Theorem8.14of[2]for:

Fact4.5Let K be a cyclic extension of F with Galois group G.Then H2(G,K?)is isomorphic to the quotient group F?/N K/F(K?).

We can now prove:

Theorem4.6(F a supersimple?eld.)Br(F)is trivial.

Proof.We will prove by induction on n,that for every?nite extension K of F and every Galois extension L of K of degree n,Br(L/K)(or equivalently H2(Gal(L/K),L?))is trivial.By the remarks above this is enough.Suppose this is proved for all n

order of G,let G p be a p-Sylow subgroup of G and let K p be the?xed?eld of G p.By induction hypothesis H2(G p,L?)is trivial,so by Fact4.4so is H2(G,L?).So we may assume that G is elementary abelian,of cardinality p say.Let K1be obtained from K by adjoining all pth roots of unity.Let L1be the compositum of L and K1.By Fact4.3it is enough to prove that H2(Gal(L1/K)is trivial.As all prime divisors of the order of Gal(K1/K) are strictly less than p it follows by the induction hypothesis,as above,that H2(Gal(K1/K),K?1)is trivial.Thus by Fact4.3,it remains to prove only that H2(Gal(L1/K1),L?1)is trivial.Note that L1is an extension of K1of degree p.

So,changing notation,we are reduced to showing that H2(Gal(L/K),L?) is trivial,when G=Gal(L/K)has order p and K,a?nite extension of F, contains all pth roots of unity.If p is the characteristic,then by Fact4.2 (ii),the restriction of N L/K to K?is already surjective,so N L/K:L?→K?is surjective,which by Fact4.5?nishes the proof.So we may assume that p is prime to the characteristic.In that case,L is a Kummer extension of K generated by a solutionαto x p=a for some a∈K1.We will show that N L/K:L?→K?is surjective.

Note that{1,α,α2,..,αp?1}is a basis for L over K,with respect to which any element of L has coordinates x1,..,x p from K.So the norm map can be represented as a map from the set of p-tuples of elements of K(not all zero)to K?.Letωbe a primitive pth root of unity.Then the conjugates ofαunder G areα,ωα,..,ωp?1α.An easy computation shows that for any x1,x2in K,N L/K(x1,x2,0,...0)=x p1+ax p2if p is not2and=x p1?ax p2if p=2.Now K,being a?nite extension of F is also supersimple.By Fact 4.2,the multiplicative subgroup T of K?of pth powers is de?nable and of ?nite index.By Proposition3.4,both T+aT and T?aT contain K?.Thus, the image of N L/K must be equal to K?.By Fact4.5,H2(L/K)=0.The proof is complete.

Note that Theorem4.6has the following consequence:

Corollary4.7Suppose that F is a supersimple?eld,and D is a?nite di-mensional division algebra over F.Then D is a?eld.

Proof.By for example15.8of[6],if a division ring is?nite-dimensional over a sub?eld then it is?nite-dimensional over its centre.

5Supersimple division rings

This section is devoted to a proof of:

Theorem5.1Any supersimple division ring is a?eld.

Corollary4.7will play a crucial role in the proof.

We proceed to the proof of Theorem5.1,which will go through various reductions and cases.

We will assume that D is a supersimple division ring(namely a division ring type-de?nable in a big modelˉM of a supersimple theory),which is not commutative,and look for a contradiction.We make continuous use of Fact 4.1.In particular SU(D)is a monomialωα.n say(αan ordinal and n a positive integer).

Now for any?nite subset A of D,the centralizer of A in D,C D(A)is a de?nable subdivision ring of D.Note that if D1

Assumption.For every a∈D\Z(D),C D(a)is a?eld.

Lemma5.2SU(Z(D))<ωαand also SU(C D(a))<ωαfor each a/∈Z(D). Proof.Otherwise,D has?nite dimension over a sub?eld,contradicting Corol-lary4.7.

Lemma5.3(i)For any a/∈Z(D)the conjugacy class of a in D,a D is generic,and moreover there are only?nitely many such conjugacy classes. (ii)Also for a/∈Z(D),{x a?x:x∈D}is an additive subgroup of?nite index of D.

Proof.(i)For a/∈Z(D),a D is in de?nable bijection with D?/C D(a)?which by Fact4.1and Lemma5.2has SU-rankωα.n.Thus a D is generic in D.As the relation of being in the same conjugacy class is an equivalence relation, there can be at most?nitely many such generic classes.

(ii)Note that C D(a)is precisely the kernel of the additive endomorphismμof D which takes x to x a?x.So by Lemma5.2,SU(Ker(μ))<ωα.Fact 4.1implies that Im(μ)has SU-rankωα.n,so has?nite index.

Lemma5.4If a,b/∈Z(D)do not commute,then C D(a)∩C D(b)=Z(D). Proof.Let c∈C D(a)∩C D(b).If c/∈Z(D)then by the Assumption above, C D(c)is a?eld,so a and b commute,contradiction.

Lemma5.5Assume that char(D)=0.Then

(i)For any b∈D\Z(D)and n>0,b n/∈Z(D)

(ii)for all a∈D\Z(D),a D?a?Z(D).

Proof.(i)Note that as char(D)=0,D has no additive subgroups of?nite index,and thus by Lemma5.3(ii),{x b?x:x∈D}=D.Thus there are nonzero c∈Z(D)and a/∈Z(D)such that a b?a=c.Then b is in the normaliser in D of C D(a)and b/∈C D(a).If b n∈Z(D)then the division ring generated by C D(a)and b is noncommutative and?nite-dimensional over the ?eld C D(a),contradicting Corollary4.7

(ii)As a/∈Z(D),(a+Z(D))∩Z(D)=?.Lemma5.3(i)says there are only ?nitely many noncentral conjugacy classes in D and so one of them,say b D, must intersect a+Z(D)in a generic subset of a+Z(D)(namely a subset of a+Z(D)of maximal SU-rank).Without loss of generality,b∈a+Z(D). Thus(b D∩(a+Z(D))?b)which equals(b D?b)∩Z(D)is a generic subset of Z(D)which is easily seen to be an additive subgroup,and thus an additive subgroup of?nite index.It follows,as we are in characteristic0that b D?b contains Z(D).In particular,note a∈b+Z(D)?b D,so also a D?a contains Z(D).

With the above lemmas we can?nish the proof.We divide into cases. Case I.char(D)=p>0.

By Lemma5.3(ii)and compactness there is a uniform bound,say m,on the index in D+of the subgroup{x a?x:x∈D}for a∈D\Z(D). We will show that every element of D?/Z(D)?has?nite order,which by Kaplansky’s Theorem(see5.15in[6])implies that D is commutative.If C D(a)has cardinality at most m then clearly a has exponent at most m in

D.Otherwise we can(by choice of m)?nd nonzero a′∈C D(a)of the form b a?b for some b∈D.Note that a and b do not commute.But a p and b do commute(b a=b+a′implies b a p=b+pa′=b).So a p is in C D(a)∩C D(b) which by Lemma5.4equals Z(D).So a has order p in D?/Z(D)?.This ?nishes the proof of Theorem5.1in the positive characteristic case.

Case II.char(D)=0.

Suppose to begin with that every element of Z(D)has a square root in Z(D). Thus,we obtain in?nitely many roots of unity{ωi:i<ω}in Z(D).Let a/∈Z(D).Then for some i

So we may assume that some c∈Z(D)has no square root in Z(D),and thus by Lemma5.5(i)c is not a square in D.Choose a∈D\Z(D),and let F =C D(a),a(supersimple)?eld containing Z(D).Let T=(F?)2,which is by supersimplicity a de?nable subgroup of F?of?nite index.Let z∈Z(D)be nonzero.By Proposition3.4there are b∈T and e∈cT,both generic over z in F such that b?e=z.Note that e is not a square in D(as c is not). As e is generic in F,e/∈Z(D)so by Lemma5.5(ii),e d?e=z for some d∈D.Thus e d=b.But b is a square,while e and thus e d is not a square, contradiction.This completes the proof of Theorem5.1,and also the paper. References

[1]G.Cherlin and S.Shelah,Superstable?elds and groups,Annals of Math.

Logic18(1980),227-270.

[2]N.Jacobson,Basic Algebra II,second edition.Freeman,1989.

[3]B.Kim,Forking in simple unstable theories,to appear in J.London.

Math.Soc.

[4]B.Kim and A.Pillay,Simple theories,Annals of Pure and Applied Logic

88(1997),149-164.

[5]E.Hrushovski,Pseudo?nite?elds and related structures,preprint1991.

[6]https://www.wendangku.net/doc/5f3800940.html,m,A?rst course in noncommutative rings,Springer-Verlag.

[7]A.J.Macintyre,Onω1-categorical theories of?elds,Fund.Math.1971.

1-25.

[8]A.Pillay,De?nability and de?nable groups in simple theories,to appear

in Journal of Symbolic Logic.

[9]A.Pillay and B.Poizat,Corps et Chirurgie,Journal of Sybolic Logic60

(1995),528-533.

[10]B.Poizat,Groupes Stables,Nur-al Mantiq wal-Ma’rifah,Villeurbanne,

1987.

[11]J.-P.Serre,Local Fields,Springer-Verlag,1979.

[12]S.Shelah,Simple unstable theories,Annals of Math.Logic19(1980),

177-203.

[13]F.O.Wagner,Groups in simple theories,preprint1997.

六年级绝对值应用(讲义及答案)精益版

绝对值应用（讲义） ? 课前预习 1. a 的相反数是_______，a b -的相反数是_______，a b c -+的相反数是 ________；若0a b c -+<，则a b c -+=________． 2. 已知0a c <<，0ab >，b c a <<，在下图数轴上标出b ，c 的大致位置． 3. 当a >0时，a =____，a a =____；当a <0时，a =____，a a =____． ? 知识点睛 1. 去绝对值： ①看_____，定_____；②依法则，留_____；③去括号，合并． 2. 分类讨论： ①_____________________________________________； ②_____________________________________________． 3. 绝对值的几何意义： a b -表示在数轴上数a 与数b 对应点之间的距离． ? 精讲精练 1. 小明得到了一个如图所示的数轴草图，他想知道一些式子的符号，请你帮他完成． a -____0，a b +____0，a b -____0，b a -____0．（填“>”、“<”或“=”） a 2. 设有理数a ，b ，c 在数轴上的对应点如图所示，则b -a ____0，a +c _____0．化简2b a c a c a -+-+-=____________． 3. 设有理数a ，b 在数轴上的对应点如图所示，化简1a b a b b +---+-． 01a b

4. 已知0a c <<，0ab >，b c a >>，化简b a b c a b c -++-++． 5. 已知0c a <<，0ab <，a c b >>，化简a a c b c b a -+----． 6. 已知0a b +<，化简13a b a b +----． 7. 若15x -=，1y =，则x y -的值为__________________． 8. 若24x +=，3y =，则x y +的值为_________________． 9. 若4a =，2b =，且a b a b +=+，则a b -的值是多少？

大亚湾规划

[编辑本段] 发展布局面对百年难遇的发展机遇和奠定百年发展格局的历史重任，经过认真思考、反复推敲，我们提出了突出大亚湾滨海特色的规划布局,力争通过5～10年时间的努力，奠定将大亚湾建设成为惠州城市南部的滨海新区，南中国海北岸的璀璨明珠的坚实基础。具体发展布局如下：（一）东区主要发展石油化学产业。按高起点、高标准规划，努力建设成为一个反映当代最新技术水平的世界级石化工业基地。（二）西区主要发展电子、汽车零部件产业，规划面积约40平方公里。根据区位及产业特点，西区可分为石化大道西段电子信息产业带、西北部东风汽车城、西南部惠深合作开发区、南部汽车电子产业区和东南部汽车零部件生产基地五大片区（带）。（三）港区以惠州港为基础，重点发展港口、物流业，总用地面积为8平方公里。凭借其天然的位置和港口优势，加上香港和记黄埔的参资入股，惠州港将进入崭新的发展时期，推动物流产业快速增长。（四）中心区中心区总用地面积约22平方公里，以行政、金融、商务、居住为发展方向，是打造沿海城区，突出滨海特色的核心区域，也是完善城市功能的重点地区。按照区位可分为北、中、南三个片区。（五）黄金海岸旅游区主要包括霞涌、辣甲岛、三门岛、五洲岛和笔架山五大景区，目标是把大亚湾区建设成集游览、观光、购物等旅游活动于一体的、接待设施先进、服务水平一流、景色环境优美的区域性旅游胜地。（六）沿海岸线重点规划生活岸线和旅游岸线，挖掘港口岸线和工业岸线的潜力，使之形成一体，共同打造出充分体现惠州沿海城市特色的滨海景观带。 [编辑本段] 开发区优势大亚湾集中展现了中国沿海地区的综合优势，具有发展成为超大规模经济体和区域经济中心的优良条件。主要体现在：

绝对值的意义及应用(最新整理)

绝对值的意义及应用绝对值是初中代数中的一个重要概念，应用较为广泛．在解与绝对值有关的问题时，首先必须弄清绝对值的意义和性质。对于数x而言，它的绝对值表示为：|x|. 一. 绝对值的实质：正实数与零的绝对值是其自身，负实数的绝对值是它的相反数，即也就是说，|x|表示数轴上坐标为x的点与原点的距离。总之，任何实数的绝对值是一个非负数，即|x|≥0，请牢牢记住这一点。二. 绝对值的几何意义：一个数的绝对值就是数轴上表示这个数的点到原点的距离。例1. 有理数a、b、c在数轴上的位置如图所示，则式子|a|+|b|+|a+b|+|b-c|化简结果为( ) A．2a+3b-c B．3b-c C．b+c D．c-b (第二届“希望杯”数学邀请赛初一试题) 解：由图形可知a＜0，c＞b＞0，且|c|＞|b|＞|a|，则a+b＞0，b-c＜0．所以原式＝-a+b+a+b-b+c＝b+c，故应选(C)．三. 绝对值的性质： 1. 有理数的绝对值是一个非负数，即|x|≥0，绝对值最小的数是零。 2. 任何有理数都有唯一的绝对值，并且任何一个有理数都不大于它的绝对值，即x≤ |x|。 3. 已知一个数的绝对值，那么它所对应的是两个互为相反数的数。 4. 若两个数的绝对值相等，则这两个数不一定相等(显然如|6|＝|-6|，但6≠-6)，只有这两个数同号，且这两个数的绝对值相等时，这两个数才相等。四. 含绝对值问题的有效处理方法 1. 运用绝对值概念。即根据题设条件或隐含条件，确定绝对值里代数式的正负，再利用绝对值定义去掉绝对值的符号进行运算。

例2. 已知：|x-2|+x-2＝0，求：(1)x+2的最大值；(2)6-x的最小值。解：∵|x-2|+x-2＝0，∴|x-2|＝-(x-2) 根据绝对值的概念，一个数的绝对值等于它的相反数时，这个数为负数或零， ∴x-2≤0，即x≤2，这表示x的最大值为2 (1)当x＝2时，x+2得最大值2+2＝4； (2)当x＝2时，6-x得最小值6-2＝4 2. 用绝对值为零时的值分段讨论．即对于含绝对值代数式的字母没有条件限制或限制不确切的，就需先求零点，再分区间定性质，最后去掉绝对值符号。例3. 已知|x-2|+x与x-2+|x|互为相反数，求x的最大值．解：由题意得(|x-2|+x)+(x-2+|x|)＝0，整理得|x-2|+|x|+2x-2＝0 令|x-2|＝0，得x＝2，令|x|＝0，得x＝0 以0，2为分界点，分为三段讨论： (1)x≥2时，原方程化为x-2+x+2x-2＝0，解得x＝1，因不在x≥2的范围内，舍去。 (2)0≤x＜2时，原方程化为2-x+x+2x-2＝0，解得x＝0 (3)x＜0时，原方程化为2-x-x+2x-2＝0，从而得x＜0 综合(1)、(2)、(3)知x≤0，所以x的最大值为0 3. 整体参与运算过程．即整体配凑，借用已知条件确定绝对值里代数式的正负，再用绝对值定义去掉绝对值符号进行运算。例4. 若|a-2|＝2-a，求a的取值范围。解：根据已知条件等式的结构特征，我们把a-2看作一个整体，那么原式变形为|a-2|＝-(a-2)，又由绝对值概念知a-2≤0，故a的取值范围是a≤2 4. 运用绝对值的几何意义．即通过观察图形确定绝对值里代数式的正负，再用绝对值定义去掉绝对值的符号进行运算．例5. 求满足关系式|x-3|-|x+1|＝4的x的取值范围．解：原式可化为|x-3|-|x-(-1)|＝4 它表示在数轴上点x到点3的距离与到点-1的距离的差为4 由图可知，小于等于-1的范围内的x的所有值都满足这一要求。

绝对值的性质及运用

知识精讲绝对值的几何意义：一个数a的绝对值就是数轴上表示数a的点与原点的距离. 绝对值的代数意义：一个正数的绝对值是它本身；一个负数的绝对值是它的相反数；0的绝对值是0. ①取绝对值也是一种运算，运算符号是“”，求一个数的绝对值，就是根据性质去掉绝对值号. ②一个正数的绝对值是它本身；一个负数的绝对值是它的相反数；0的绝对值是0. ③绝对值具有非负性，取绝对值的结果总是正数或0. ④任何一个有理数都是由两部分组成：符号和它的绝对值，如：5-符号是负号，绝对值是5. 求字母a的绝对值： ① (0) 0(0) (0) a a a a a a > ? ? == ? ?-< ? ②(0) (0) a a a a a ≥ ? =? -< ? ③(0) (0) a a a a a > ? =? -≤ ? 利用绝对值比较两个负有理数的大小：两个负数，绝对值大的反而小. 绝对值非负性：如果若干个非负数的和为0，那么这若干个非负数都必为0. 例如：若0 a b c ++=，则0 a=，0 b=，0 c= 绝对值的其它重要性质：（1）任何一个数的绝对值都不小于这个数，也不小于这个数的相反数，即a a ≥，且a a ≥-；（2）若a b =，则a b =或a b =-；（3）ab a b =?； a a b b =(0) b≠；（4）222 |||| a a a ==； a的几何意义：在数轴上，表示这个数的点离开原点的距离． a b -的几何意义：在数轴上，表示数a．b对应数轴上两点间的距离．【例题精讲】模块一、绝对值的性质【例1】到数轴原点的距离是2的点表示的数是（） A．±2 B．2 C．-2 D．4 绝对值

大亚湾的区域优势

大亚湾中心区与西区的优势中心区的优势：随着中国人口的增长，人们对于住房的需求也越来越强烈，特别是今年，政策的调控，房价的上涨和租金的上涨，越来越多的人选择自己买房，而不在租房，有钱人也选择买房来投资，都知道人民币在贬值，炒股十年一场空，买房十年富如翁，事实也证明确实如此。作为离着深圳最近的惠州大亚湾，之前一直不被看好，但随着厦深高铁沿海高速的开通和惠大高速，惠州机场还有规划中的地铁，逐步的开始运行和世界500强企业的进驻（比亚迪，中海壳牌等），越来越多的深圳客选择东进，来惠州大亚湾买房子了，大亚湾地大楼多，一些比较不熟悉大亚湾的深圳人，一看到那么多的选择，肯定会觉得茫然，那我们作为大亚湾的专业售楼经纪人，应该引导客户在我们所售的楼盘购房，让客户认可我们所售片区的优势，选择适合客户的房子。大亚湾就行政规划而言主要分为五大片区，西区，中心区，澳头老城区，石化区，黄金海岸度假区。首先介绍一下我们大亚湾的中心区，中心区是CBD的高档居住区，主要以行政，商业，居住，金融为主的行政中心地带，大亚湾这座位于广东省惠州市占地20多平方公里的沿海新城，拥有坐拥三面环山，一面靠海的优异环境，在政府规划后，中心区便成为投资，居住，养老最适合的地方，也是政府重点发展的片区，因为它是集学校，医院，市场，商业为一体的一个高档生活区。学校：成长树幼儿园，晶晶幼儿园，岩前小学，黄鱼涌小学，还有正在规划要建设的二小，澳头中学等，住在中心区的业主买房可享受12年义务教育，全部公立学校，住在中心区任何楼盘到达学校均在5分钟车程，可节约很多时间出来，而且学校都有校车接送，高端教育质量，所以，在中心区买房孩子上学无忧。医院：中心区还拥有占地120亩的中山大学附属医院，惠亚医院由政府全资兴建，按照三级甲等标准建设，是一所集医疗，教学，科研为一体的公立费盈利综合医院，拥有现代化的标准医疗设备，您居住养老无忧，车程仅需4分钟。公园：红树林公园占地5.1万平方米，东起白寿湾大桥，西至中兴二路，全长3.9公里拟打造一心两翼十景格局，“一心”已建成的红树林公园为核心区域，以中兴中路为连接点展开城市活力翼，“十景”打造虎头山远眺，龙海望峰，兰州轻渡，长提栖鹭等做十景。红树林以生态保护，科普教育，休闲为功能的湿地公园，是中心业主，早晚散步，休闲，跑步的好地方，这里有栖息枝头的白鹭，有日落的夕阳，是居住养老的圣地。

惠州大亚湾猴仔湾项目一期规划设计说明书

目录目录 ......................................................... 1 1.现状情况 ................................................... 1 区域位置 ................................................... 1 自然条件 ................................................... 1 用地及建设现状 ............................................. 1 道路交通现状 ............................................... 1 景观现状 ................................................... 1 2.规划目标、原则和依据 . (1) 规划目标 ................................................... 1 规划原则 ................................................... 1 规划依据 (2) 3.总体布局 (2) 周边规划 ................................................... 2 用地功能布局 .. (2) 4.道路交通规划 (3) 现状概括 ................................................... 3 规划原则 ................................................... 3 道路系统规划 ............................................... 3 交通设施规划 .. (3) 5.用地竖向规划 (3) 规划目标 ................................................... 3 规划原则 ................................................... 3 竖向规划控制 .. (4) 6.绿化设计 (4) 现状概况 ................................................... 4 规划原则 ................................................... 4 规划目标 ................................................... 4 绿化设计 (4) 7.市政综合管线规划 ............................................ 5 供电工程规划 (5) 通信工程规划 ...................给水工程规划 ...................污水工程规划 ................... 雨水工程规划 ...................防洪排洪规划 ...................燃气工程规划 ...................综合管线规划 ...................

绝对值几何意义应用

仅供参考学习个人收集整理绝对值几何意义应用一、几何意义类型：0a?a?a 类型一、0：表示数轴上地点地距离；到原点 ab??b?a bb aa 地距离（或点；类型二、：表示数轴上地点到点到点地距离） )?baa?b??()?ab?(?b?b aa?地距离）：表示数轴上地点到点类型三、到点；地距离（点ax?ax ：表示数轴上地点地距离；到点类型四、)a?(?x?a?x xa?. 类型五、到点：表示数轴上地点地距离二、例题应用：4?xx?4x ，则、地几何意义是数轴上表示地点与表示地点之间地距离，若例1.（1）=2?x. 3x?1?x?3x ，则（2）地几何意义是数轴上表示地点与表示地点之间地距离，若、?x. 15??qm若3）、如图所示数轴上四个点地位置关系，且它们表示地数分别为m、n、p、q.，（1 n?q?n，pp?m?，15m???m?8np??q?n?1，qp3 ;若，则，n?p?.则 a?b?b?c?a?cc，，ba，，如果在数轴上地对应点为A，（4）、不相等地有理数B，C. 在数轴上地位置关系B，，C 则点A a?b?9，c?d?16且a?b?c?d?25da、cb、、，求均为有理数，拓展：已知 b?a?d?c的值. ??且a?b?c?d?25.25a?b?c?a (9b?)??16?ddc???解析： ?b?9?a，c?d?16?b?a?d?c?9?16??7. 3x??x?32?x?x时，取最大值，最大）（例2.1、①当取最小值；②时，当值为. 1 / 8 个人收集整理仅供参考学习 x?3?x?2?7x?; 利用绝对值在数轴上地几何意义得（2）、①已知，

惠州市惠城区南部新城控制性详细规划

惠州市惠城区南部新城规划道路交通规划本片区对外联系的主要通道为惠大高速公路、莞惠高速公路、惠南大道、主1号路、主3号路（三环路南延段），各建设地块出入口及内部支路详见图则。本片区内道路采用网状道路结构，区内道路分为六个等级：（1）高速公路：惠大高速、莞惠高速，红线宽度控制60－80米；（2）快速路：惠南大道、四环路，红线宽度60米；（3）城市主干路：主1号路、主3号路（三环路南延段）、主2号路、主4号路、6号路，红线宽36－60米，双向6～8车道；（4）城市次干路：红线宽32—42米，双向4～6车道；（5）Ι级支路：联系各街坊及组团间的支路，红线宽18—32米；（6）Π级支路：各街坊及组团内部道路，红线宽18米。南部新城的规划填补了惠城区和惠阳区两个市区主要城区过渡地带城市规划空白的现状，南部新城的规划区域，既是惠城区、惠阳区两大主城区城市中心的延伸，也是相对较为独立的一个崭新城市区域。南部新城具有良好的区位条件，是连接各城区的重要交汇点《惠州市惠城区南部新城控制性详细规划》指出，南部新城具有良好的区位条件：广东省区域空间的资源结构性变化，使得惠州市成为区域发展的重点地区之一。在中海壳牌南海石化项目建成投产、大亚湾经济技术开发区发展迅猛的条件下，惠州市域经济发展重点正向南调整，这也使得南部新城成为惠城区及惠阳区（包括大亚湾经济技术开发区），东西向连接陈江仲恺地区惠东的重要节点。在惠州城市总体规划平面图中，南部新城正好处在惠州的中心位置。南部新城规划是市政府历年来首次对惠城区和惠阳区的中间地带进行详细规划。惠城区和惠阳区两大主城区存在很多优势互补的方面，两个主城区的发展呈现出“向心力”。南部新城规划的出台，将有利于整个惠州城市资源的整合和升级。此外，南部新城将成为惠州城市发展南北走廊的重要衔接点，是惠州城市“南进北拓”的重要跳板。根据惠州城市总体规划，惠州主要由四大城市次区域组成，即分为惠城、惠阳-大亚湾、陈江-仲恺和北部山区四个主要组成部分。新规划的南部新城，属于惠城次区域的一部分。南部新城的发展，将承担起疏导老城区城市功能的作用，包括居住、城市商务和公共服务等。优越的自然资源，将为南部新城乃至整个中心市区奉献出最美的人居环境驱车从演达一路向南，片刻间进入惠南大道，地势豁然开朗，视野渐觉舒展，青山绿水之间，金山片区的原生态身姿就展现无遗。其实，不少人对南部新城的自然生态了解不多，这里的一切仍处在浑沌未开、朦胧面纱依在的初始开发阶段。单单以南部新城最受瞩目的金山湖为例，对其了解较深的人也不是太多。当年惠州学院选址从木墩湖

绝对值应用(绝对值的几何意义)(北师版)(含答案)

学生做题前请先回答以下问题问题1：绝对值的几何意义： ①表示在数轴上，x所对应的点与_______的距离． ②表示在数轴上____________________________对应点之间的距离． ③表示____________________________对应点之间的距离．绝对值应用（绝对值的几何意义）（北师版）一、单选题(共10道，每道10分) 1.已知，则a，b的值分别为( ) A.a=3，b=5 B.a=-3，b=5 C.a=3，b=-5 D.a=-3，b=-5 答案：B 解题思路：试题难度：三颗星知识点：绝对值的非负性 2.若，则ab=( )

A.0 B.3 C.-3 D.±3 答案：C 解题思路：试题难度：三颗星知识点：绝对值的非负性 3.若与互为相反数，则a+b=( ) A.-1 B.1 C.5 D.-5 答案：A 解题思路：试题难度：三颗星知识点：绝对值的非负性 4.若x为有理数，则的最小值为( )

C.3 D.5 答案：A 解题思路：试题难度：三颗星知识点：绝对值的几何意义 5.若x为有理数，则的最小值为( ) A.1 B.3

答案：D 解题思路：试题难度：三颗星知识点：绝对值的几何意义 6.若x为有理数，则的最小值为( ) A.1 B.2 C.3 D.4 答案：B 解题思路：

试题难度：三颗星知识点：绝对值的几何意义

7.若x为有理数，则的最小值为( ) A.2 B.3 C.4 D.5 答案：C 解题思路：试题难度：三颗星知识点：绝对值的几何意义 8.当x=____时，有最_____值，是_____．( ) A.0，小，6 B.0，大，6 C.0，小，0 D.0，大，0 答案：A 解题思路：试题难度：三颗星知识点：利用绝对值的非负性求最值 9.当x=____时，有最_____值，是_____．( ) A.4，小，3 B.4，大，-3 C.4，小，-3 D.0，大，3 答案：C

惠州大亚湾经济技术开发区详细介绍

大亚湾经济技术开发区惠州市大亚湾经济技术开发区于1993年5月经国务院批准成立，面积为9.98平方公里，2006年3月经国务院批准扩大到23.6平方公里。包括经济技术开发区在内的大亚湾规划区于1991年6月由广东省人民政府批准设立，辖陆地面积265平方公里，海域面积1300平方公里。面对百年难遇的发展机遇和奠定百年发展格局的历史重任，经过认真思考、反复推敲，我们提出了突出大亚湾滨海特色的规划布局,力争通过5～10年时间的努力，奠定将大亚湾建设成为惠州城市开发区布局图南部的滨海新区，南中国海北岸的璀璨明珠的坚实基础。具体发展布局如下：（一）东区主要发展石油化学产业。按高起点、高标准规划，努力建设成为一个反映当代最新技术水平的世界级石化工业基地。（二）西区主要发展电子、汽车零部件产业，规划面积约40平方公里。根据区位及产业特点，西区可分为石化大道西段电子信息产业带、西北部东风汽车城、西南部惠深合作开发区、南部汽车电子产业区和东南部汽车零部件生产基地五大片区（带）。（三）港区以惠州港为基础，重点发展港口、物流业，总用地面积为8平方公里。凭借其天然的位置和港口优势，加上香港和记黄埔的参资入股，惠州港将进入崭新的发展时期，推动物流产业快速增长。（四）中心区中心区总用地面积约22平方公里，以行政、金融、商务、居住为发展方向，是打造沿海城区，突出滨海特色的核心区域，也是完善城市功能的重点地区。按照区位可分为北、中、南三个片区。（五）黄金海岸旅游区主要包括霞涌、辣甲岛、三门岛、五洲岛和笔架山五大景区，目标是把大亚湾区

建设成集游览、观光、购物等旅游活动于一体的、接待设施先进、服务水平一流、景色环境优美的区域性旅游胜地。（六）沿海岸线重点规划生活岸线和旅游岸线，挖掘港口岸线和工业岸线的潜力，使之形成一体，共同打造出充分体现惠州沿海城市特色的滨海景观带。 (一)经济发展渐入佳境良好的投资环境，吸引了来自荷兰、美国、英国、日本、新加坡、德国等20多个国家和地区的客商前来投资，其中有世界500强企业12家。2006年，全区实现地区生产总值100.2亿元，增开发区地图长90.3%；实现工业总产值284.56亿元，增长261.7%，工业销售产值273.53亿元，增长253%；惠州港吞吐量1639.2万吨，增长29.9%；国税、地税两税总量76.7亿元，增长32.56%，地方财政一般预算收入4.34亿元，增长35.83%；实际利用外资2.04亿美元，下降47.6%（下降原因是中海壳牌石化联合工厂已基本完成投资。若剔除该因素，同比增长296.4%）；完成外贸出口5.81亿美元，增长47.3%。（二）石化工业区建设快速推进惠州大亚湾石化工业区规划总面积29.8平方公里，已被广东省政府列为五个重点发展的石油化工基地之一。并于2005年4月被中国石油和化工协会授予“中国石油化学工业（大亚湾）园区”牌匾。目前，石化区内已落户项目共28个，总投资额逾730亿元；在谈的石化中下游项目11个，投资额超过80亿元。其中，中海壳牌南海石化联合工厂建成投产；中国海油惠州炼油项目累计完成总进度11.2%，项目总体设计、工艺技术选择和工艺包装设计工作已完成；惠州液化天然气电厂项目1、2号机组已并网发电；华德石化项目已完成一期工程2座10万立方米原油罐及相应配套设施的建设。（三）汽车零部件及电子信息产业形势喜人汽车零部件、电子信息、钢铁等产业形势喜人，主要集中在开发区的西部。目前，

惠州环大亚湾新区产业发展专项规划

《惠州环大亚湾新区产业发展专项规划（2014—2030）》公示 2020年建成世界级石化产业基地惠州环大亚湾新区产业发展专项规划公示以发展石油化工、海洋科技、电子信息、装备制造、商贸物流和高端旅游为重点，建成特色现代产业体系；石化产业规模位居全球前列，成为世界级生态型石化产业基地；战略性新兴产业、先进制造业、高端海洋产业比重不断上升，建成珠三角现代化高端制造基地；以滨海旅游和生产性服务业为主导的现代服务业竞争力不断提升，成为珠江口东岸现代服务业功能区和粤港澳滨海休闲度假旅游区。自6月27日惠州环大亚湾新区党工委、管委会揭牌成立以来，环大亚湾新区未来如何发展备受关注。昨日，《惠州环大亚湾新区产业发展专项规划（2014—2030）》公示。

近期→2017年目标稔平国际旅游半岛建设基本完善根据《规划》，到2017年，环大亚湾新区石化产业转型升级步伐加快，世界级生态型石化产业基地初步形成。“稔平国际旅游半岛”建设基本完善，在国内具有较高知名度。引进2～3家大型物流企业，打造一批重点物流园区，惠阳成为环大亚湾新区的现代商贸中心，现代服务业体系基本完善，辐射带动能力显著增强。陆岛资源一体化开发取得成效，海洋经济发展加速。农业四大特色产业规模进一步扩大，“三带四区”的农业发展新格局基本形成。实现工业总产值5200亿元，战略性新兴产业增加值占生产总值比重达12%，高新技术产品产值占规模以上工业产值的比重达42%，三次产业比例调整为3︰68︰29。中期→2020年实现工业总产值8500亿元根据《规划》，到2020年，现代产业体系完善，现代服务业、战略新兴产业比重显著上升，形成以先进制造业、战略性新兴产业和现代服务业为主的产业结构，海洋经济规模不断增大，特色农业发展水平明显提升。基本建成世界级石化产业基地和珠三角高端制造基地，粤港澳滨海休闲度假旅游区和珠江口东岸现代服务业功能区的地位基本确立。

浅谈惠州大亚湾石化基地区位分析

浅谈惠州大亚湾石化基地区位分析【摘要】：石油化工产业作为惠州一个重要产业部门之一，已经成为了带动惠州经济发展的一个重要环节。本文通过调查分析石化产业分布的区位条件，分析得出惠州大亚湾石化基地能够吸引石油化工相关企业在此落户的区位优势。通过查阅相关资料与调查分析，在惠州影响石化企业落户的区位因子中，市场与交通是其中最主要的两个方面，同时政策也对石化企业的落户有重要影响。综述全文，要更好促进惠州大亚湾石化基地的发展，应该完善交通网，加强对外联系，以扩大经济腹地，从而强化市场优势。同时，政府也应出台优惠政策，加强管理，营造优良创业与市场环境。【关键词】：大亚湾石化基地石化产业企业区位条件 1 引言：随着惠州近十年来的发展，惠州大亚湾石化基地已初具世界规模，已成为广东四大石化基地之一，并且惠州大亚湾石化基地发展前景广阔，有望成为像荷兰鹿特丹那样的世界级石化基地。本文通过查阅相关文献与资料，分析得出在惠州大亚湾石化基地发展的过程中，惠州许多方面的区位条件都发挥着重要的作用。惠州优越的地理位置及良好的自然条件；惠州地处珠三角工业区与毗邻香港的优越市场条件；由京九铁路与广梅汕铁路构成的铁路网、惠盐、深汕、广惠、惠河4条高速公路及一些构成的公路网和惠州港为基地的发展提供的便捷海陆交通；中外石化龙头企业在惠的投资提供的资金支持及众多相关企业在大亚湾的集聚所形成的规模经济效益；惠州政府制定的一系列优惠政策与不断完善的基础设施形成的优良产业环境等。

2 正文：一、研究背景（一）、惠州市地理概况：惠州市位于广东省东南部，珠江三角洲的东北端，南邻南海大亚湾并毗邻香港与深圳，北连河源市，东接汕尾市，西邻东莞市和广州市郊区。惠州市属亚热带季风气候，北回归线从博罗县的杨村和龙门县的南昆山穿越，东江和西枝江横贯市中部。境内北部多山地，中部和沿江地多冲积平原，阳光充足，气候温和，年平均降雨量1700毫米左右，年平均气温22℃左右。广东惠州，地处珠三角，是广东乃至全国的经济快速兴起城市之一。惠州区域优势得天独厚，素有“粤东门户”之称，地处通衢要冲，是衔接粤东和内陆省区并连接港澳台及东南亚的交通枢纽。距香港陆路80多公里，水路47公里，是广东境内除深圳外距离香港最近的中等城市。惠州具有亿吨开发潜力，紧靠国际航线，经批准对外国籍船舶开放的深水良港—惠州港。从港口修筑陈澳铁路与京九铁路接轨，是京九铁路南端便利的出海口[1]。十年前，惠州只是南粤一座小城；2001年以来，惠州依靠毗邻港澳的独特地域优势，以及紧紧把握中国加入WTO带来的一系列机遇，加大产业集群的发展力度。如今，惠州已经形成石化、电子、灯饰、汽车零部件、制鞋、纺织、服装等六大产业集群，其中大亚湾石化基地已经达到世界规模和水平。（二）、石油化学工业简介石油化学工业简称石油化工，石油化学工业是基础性产业，它为农业、能源、交通、机械、电子、纺织、轻工、建筑、建材等工农业和人民日常生活提供配套和服务，在国民经济中占有举足轻重的地位。是化学工业的重要组成部分，在国民经济的发展中有重要作用，是我国的支柱产业部门之一。石油化工指以石油和天然气为原料，生产石油产品和石油化工产品的加工工业。石油产品又称油品，主要包括各种燃料油（汽油、煤油、柴油等）和润滑油以及液化石油气、石油焦

绝对值几何意义应用

绝对值几何意义应用一、几何意义类型：类型一、0-=a a ：表示数轴上的点a 到原点0的距离；类型二、 a b b a -=-：表示数轴上的点a 到点b 的距离（或点b 到点a 的距离）；类型三、)(b a b a --=+)(a b --=：表示数轴上的点a 到点b -的距离（点b 到点a -的距离）；类型四、a x -：表示数轴上的点x 到点a 的距离；类型五、)(a x a x --=+：表示数轴上的点x 到点a -的距离. 二、例题应用：例1.（1）、4-x 的几何意义是数轴上表示x 的点与表示的点之间的距离，若4-x =2，则 = x . （2）、3+x 的几何意义是数轴上表示x 的点与表示的点之间的距离，若13=+x ，则 = x . （3）、如图所示数轴上四个点的位置关系，且它们表示的数分别为m 、n 、p 、q.若15=-q m ， 8 10=-=-m p n q ，，则=-p n ;若15=-q m ，，，q n n p m p -= -=-3 1 8 则=-p n .

的几何意义得; ③已知4 + -x x，利用绝对值在数轴上 + 3= 2 的几何意义得; 拓展：若8 1 +a a，则整数a的个数是 - + 2= 2 7 4 . ④当x满足条件时，利用绝对值在数轴上的几何意义2 3+ -x x取得最小值， + 这个最小值是. 由上题③图可知，5 +x + x，故而当 - 3 2≥≤ -x时，最小值是5. 3 2≤ ⑤若a -2 + 3时，探究a为何值，方程有 x= x +

解？无实数解？档案：5≥a ；a <5. 特别要注意的是：当x 在32≤≤-x 这个范围内任取一个数时，都有523=++-x x . 例题拓展：①若23++-x x >a 恒成立，则a 满足什么条件？答案：a <5. ②若23++-x x a 恒成立，则a 满足什么条件？答案：a ＜5-. 由上图当x ≤2-时， 2 3+--x x 5=；当x ≥3时， 23+--x x 5 -=；当2-＜x ＜3， 5 -＜23+--x x ＜5，所以5-≤23+--x x ≤5.则a ＜5-. ④若23+--x x 5. 拓展应用：已知()()()36131221=++-++--++z z y y x x ，求z y x 32++