xxx1(x

x,

1（也稱基本不等式）即(xx...x) .僅當(dāng)所有x1 n2時(shí)

1(xx )20.xx n3時(shí)

1(xxx) (3x)3(3

)3(3x)333x3x3

0.

x1x2x3時(shí)取到.對(duì)n

aba

ab2a2a2a22b2b2b22abR,i1 2

n

i等號(hào)當(dāng)且僅當(dāng)a1a2an0或bikai時(shí)成立（k為常數(shù)，i1,2n）現(xiàn)將它的證

f(x)axb2axb2axb x2ab x2ababn2a xbb nn1 2n n

a2a2 ana2a2 ann4ababa 4aa 2 bb n n1 2na aana aan2 bb nn當(dāng)且僅當(dāng)aixbix0i1

a2 ana即a即n1．(2011復(fù)旦)設(shè)nx

(x0)的最小值 xx xx x1

(n

n1.nx1時(shí)成立2．(2004復(fù)旦)比較log2425和log2526的大小解:作差,換底（使底數(shù)相同lg25 lg26 lg225lg24lg26log2425-log2526 須比較lg225與lg24lg26大小.利用均值不等式lg24lg lg24lglg24lg 從而lg24lg26lg225.從而 a,b,c

b

的最小值ab c 解:原式cab c b

1c“”僅當(dāng) b+c即ac,b0時(shí)取到c b 設(shè)x02x11x)2的最小值11(1x)1(1+x)21x1(1x)1(1+x)21x21x 2(1x)1xx0”取到xyz,u滿足2xxyzyzu1x2y2z2u的最大值42x2y2解:2xxy42x2y2442x2y2z2u14

x2y2z2u

1x1y2z1u1時(shí)取到 設(shè)實(shí)數(shù)a,b,c,d,e滿足abcde8a2b2c2d2e216,e的最大值. abcd8a2b2c2d216由柯西不等式,(12121212a2b2c2d2abcd)2e[0,得4(16e2)(8e)2解 當(dāng)abcde 時(shí)5

165 2

設(shè)正數(shù)a,b,c,d滿足cdab)

， 的最小值(a2b2(a2b2)(a2b2(a2b2)(c2d2解:(acbd)

b3d

2b2)2 (ac (acacbd. ”當(dāng)abcd

時(shí)取到2給定正整數(shù)k和n(k<n).已知給定正實(shí)數(shù)a1a2ak.試求正實(shí)數(shù)ak+1ak+2an 1Ma1a2anaa an取最小值 na,b11a,b11k 1 akM(a

1a2a)b11k 1aa)b11k 1akabanab

,即

k1

時(shí),

min

nk)29．(2008同濟(jì))x2y2z244解:(x2y2z28)26224)28x6y249262392,即柯西不等式等號(hào)成立,由其條件知4xxy

解得y z(2010浙大

1的正數(shù)x1x2xn滿足x1x2xn1

... 4x

x

x 1:由柯西不等式(xx3)(xx3) (xx3)

...

n2

x x xx3注意到(xx3xx3

n x3xx x1)n1(x3x3 x3(0,1(因xx3xx x1)n 1從 1x

...x

n222, ,2:可先證一個(gè)局部不等式

x3i x3i

(*)(i1, (*)14x24x4(2x21)2 xn)n式相加得原不等式左邊4(x1xn)

顯然成立4”成立時(shí)

xi

2i1 n.這顯然不可能2a,b,c滿足abc1(a1)2(b1)2(c1)2100

左邊(a2b2c2

1

1c2(x(xx3)(xx x)(x3x3n x3n33a2b2c2

336(”在abc33333a2b2c233a2b2c22732732

33a2b2c233a2b2c21(3a2b2c22732733

(3abcabc1 2080100(”在abc1時(shí)取 2.由柯西不等式1

1

1 1左邊3ab c 3 c1

1

(因(abc

933

abc

ca,b,c1(a11)(b11)(c11 1:令axbyczxyz可由abc解出 其中,x,yzR,可把原不等式齊次化為(xyz)(yzx)(zxy)xyz(*)(c

1uxyz,vyzx,wzxy,三數(shù)兩兩之和為正,至多一者非正.若恰有一數(shù)非正,則另兩數(shù)為正,uvw0xyz,(*)成立.若u,v,wR,由均值不等式

12xx(xy(xyz)(yz類似

于是uvw

(*)成立

x21(abbb 解2:設(shè)a x,b y,c z,則 y2b 1bab z2c111(abb 于是

x y z整理,得4xyyzzxxyz又xyyzzx33(xyz)23t2,(記t3xyzx2x15于是x2x15

(t1)(t2)2

故t

設(shè)x

[,2

,證明： x解: 153x)2(211)(2(x1)(2x3)x4(14x419.(因x5故原不等式左邊

.由柯西不等式成立條件知兩個(gè)等號(hào)無法同時(shí)取到設(shè)a1a2an1,2,…,n的一個(gè)排列.1a1

a2

...

an1

nn 解:(aa)(aa) a)

左邊 2(a1a2 an)a1 n(n1)a1

n(n1)1 (n1)(n2) (n1)(n n利用均值不等式證明：數(shù)列:解:①證:

11)n單調(diào)遞增且有上界n

1

nn

n11 1

1

1 1

n 因n1

1 1 1 n

n1

n n n

n n知an1an②證an有上界定義

(11)n1,可證 (須比較b與

1)n2

n

)n1 )n2

)n n n1 n

1)n1

n2 n1 n1 (n2)n n1n1n n n而n

n1

n(n1)21

n(n

n n n(n(n

(n1)2 anbnbn1 b14.4是an的一個(gè)上界n注:高等數(shù)學(xué)有結(jié)論:單調(diào)有界數(shù)列必有極限,故lima存在,我們記為na,b,c2a+3b+4c=22239的最小值 設(shè)x,y,z11112 xyxyxyxy對(duì)滿足1rst的一切實(shí)數(shù)r,s,t,求wr1)2s1)2t1)241)2 值5已知實(shí)數(shù)a,b,c滿足a2abacbc65(2008浙大)a,b

,3a+b+2c的最小值n(ab)(a2n122n(ab)(a2n122a a a比較2nn2n1,2112

n2,2222

n3,2332n4,2442;n5,25nk5時(shí)2kk2nk1時(shí)2k122kk211)kk22k1k6n，每一個(gè)正方形都可以分成n個(gè)正方形nn

nnk(k6若成立,可分k3余數(shù),并利用“奠基”中的構(gòu)造,形為四個(gè)357的正整數(shù)n分的郵資n8,nk時(shí)k7)成立①3分支付,3張.25分即付k1②5分,23分設(shè)有2n(n1m1m2，則從甲堆拿m2個(gè)球放到乙堆中去，這樣n1時(shí),221nk2k2k1個(gè)球分成若干堆,注意到總球數(shù)是22除后余數(shù)(即奇偶)分兩類,顯然奇數(shù)個(gè)數(shù)的堆共有偶數(shù)堆,將它們兩堆分一組,移動(dòng)后可都變成偶數(shù)個(gè)球的堆.于是每堆都有偶數(shù)個(gè)球.每堆中的球配成2球“球包”,求證：11... 證:n1234命題成立n5時(shí),加強(qiáng)證1

...

3 nn5時(shí)左邊1.761,右邊1.775.即n5nnk5時(shí)成立,1

...

3 kknk1kk

1 1(k1(k 3k3 3 k (k (k 3k2作差比較3 3 k2

(k1)3 3k(k3k(kk(3k2)k23(k(k k9(k9(k1)33k2k即nk1時(shí),1

9(k9(kk

01(kk1(kk綜上,1

...

3(nN*)成立64片金片，這就是所2641)秒n1f(n211顯然若nk時(shí) f(k)2k1,則對(duì)于nk1,想要全部移走nk1個(gè)金片(如從A3步:①將上層k片移到C,B,③將C B.f(k12f(ka(a2

1且

，求證：數(shù)列{an3a2 n

a(a2 2a(1a)(1a證:顯然{an}是正項(xiàng)數(shù)列.且

a3a2

n3a2 所以anan10an1anan1an0a11,可歸納0an1

n1時(shí)成立;nk時(shí)成立,nk1時(shí)a(a2 a(a2 (1ak akk

k 0,3a21

1ak

1k 0,3a21 3a21

ak1類似可證a11時(shí),an已知數(shù)列

滿足

1n，有121

2

a1,a1,a1,a1

猜想an2n11再歸納,即可nx

,...xx

...x

.).2

2n1時(shí)成立;若nk時(shí)成立,即(1x)(1x)...(1x1 nk1時(shí),不妨令

x

,則由歸納假設(shè)可知(1x)(1x)...(1y1 k

1又因?yàn)?y1x1

(1x)(1

),所以1x)(1x)...(1y k

k

1x1)(1x2)...(1xk1)(1xk)(1xk1,nk1時(shí)命題成立1222，求證：112

...1nnkn2時(shí)顯然成立nnk12k若nk時(shí)成立,即 ...12k

,則當(dāng)nk1時(shí) 2kk左邊 ... k1 2kk

k1 kk

k

kkkkk k1 k1 k1 k1 k

k1,nk1”成立

}n

a2a2a2n完全平方數(shù)

證:a2a2a2是奇數(shù)的完全平方數(shù)”.當(dāng)n1時(shí),a5即可 設(shè)nk時(shí)結(jié)論成立,即存在k個(gè)正數(shù)aa a使a2a2...a2為奇數(shù) 2m1的平方,即a2a2...a2(2m1)2.取 2m22m,

k1(2m1)22m22m)22m22m1)2也是奇數(shù)的平方.又1

a2a2...a21a21

.有 a.于是nk1時(shí)結(jié)論也k

k 1 )首項(xiàng)為正數(shù)的數(shù)列{aa=a2+3),nNn n4 4

4

n=kak是奇數(shù).n=k+1,ak+1是奇數(shù),n=k+1(2)由a1a2+3)>aa2-4a3>0解得a>30<a2

x2

n=k+10<ak+1<1,n=k+1a1<3a2>3n=2則a-a1a2+3)-a1(a1)(a3)>00<a<1a>3nN*,n n4

4

121

3

n(2n1)(2n

n(n1)2(2n1)nk1時(shí),左

k(k1) (k (k1)(k2(2k (2k1)(2k 2(2k求證：112

2n112 k 12 k 1k1 k1 k1

1k2k

k 2k n11n

n

...11k1k21k1k2 (k21k

k2k1(2k

k2

k(k2{a

a5

a

a

n滿足5

,

1)，求證：n1時(shí)

a122,nk時(shí)不等式成立,即ak2(k (a則 2 2k 0,a 2,k

2(a

k 0x

x1

x

x若

且

，求證：n證:加強(qiáng)命題:對(duì)n

*,1

1x

x成立xn①n1,0x11x1n

11

,.1②若nk時(shí)成立,即1xk1x

則當(dāng)nk1時(shí)

xk1xkx

x1xx1xxk11xk

x1x

1

.成立S2010N為滿足0N22010S的子集進(jìn)行黑(3)N個(gè)白子集證:S是一個(gè)n元集合,①當(dāng)n1時(shí),S的子集為S，.對(duì)于0N

件;nk時(shí)結(jié)論成立,那么對(duì)于nk1

Sa1a2 akak1}.S ,A2k,將S的含ak1的子集分別記為B1,B2 ,B2k,則 N2k,用歸納假設(shè)將A,A ,A中的N個(gè)子集染成白色,其余染黑 (1),(2).然后將B1,B2 ,B2k全染黑,這種染色方法恰有N個(gè)白子集,且任意兩個(gè)白子的并集仍為白子集,任意兩個(gè)黑子集的并集仍為黑子集.若2k1N2k1, N2kr,1r2k,則用歸納假設(shè)將A,A ,A中的r個(gè)染成白,另2kr 使其滿足條件(1)和(2),再將B1,B2 ,B2k全染白,仍可滿足.綜上

n2010可成立

f(x)在閉區(qū)間[a,b]f(a)f(b)<0，那么在開區(qū)間(a,b)c，使f(c)=0整系數(shù)多項(xiàng)式若有有理根，q/p（p、qp，qnn1、(2012復(fù)旦)x3pxq03個(gè)根互不相同，且可成等比數(shù)列，則它們 解:xqxq'2x xq'xq'2x01q'q'20,q'1 3 x2、(2012北約)解方程：x

3|

x116xxx116xx2710xx

x2知1|t3||t5||(t3)(t5)|2,.知原方程無實(shí)根3(2011復(fù)旦)設(shè)abb0，、、x3axb0、1 1、解:由韋達(dá)定理

a1111112()2a

1111()()

1

1

2

2 2

1 1 1 ()()( ()()( 而

22

22

可得方程:b2x32abx2a2xb0474、(2005復(fù)旦)在實(shí)數(shù)范圍內(nèi)解方程410x47uv解:換元,令u410xv47

則 uv u4v4(u2v2)22u2v2((uv)22uv)22u2v28136uv2u2v2uv

u u得uv2或16(舍

聯(lián)立uv

得v2或v1,x6或 5、(2005交大

x3ax2bxc0的三根分別為abc,abc解:由韋達(dá)定理

a(abbabbcca,c0或ab1c2ab

a

a若c0,代入①,②得 得b0(舍)或b2b

c

c2abcab1,代入①,②得bbcca

c(2ab)代入⑤得bab)(2ab12a23abb21a1代入得b2

b22

b4b32b220b注意到1是解們都不是

(b1)(b32b20.對(duì)b32b20,若有有理根,必為12,a a綜上

abc為b1b2c

c6f(x0)x0x0f(xff(xf(xf(xff(x證:(1)x0ff(x的唯一不動(dòng)點(diǎn),ff(x0x0,f(x0

tf(x點(diǎn),f(tx0ff(tf(x0t,tff(x的不動(dòng)點(diǎn),ff(xx0

f(t

tf(x的不動(dòng)點(diǎn).存在性得證!唯一性:若有ttf(t')t',則f(f(t'))f(t')t'.t'是f(f(x))異于t的不動(dòng)點(diǎn),(2)abf(x的兩個(gè)不動(dòng)點(diǎn)ab).ff(x的不動(dòng)點(diǎn),ff(x有第三個(gè)不動(dòng)點(diǎn)t(atb且唯一(即ab,t是唯一的三個(gè)).f(f(t))t.則f(f(f

fff(tf

f(tff(x的不動(dòng)點(diǎn)!ff(x三個(gè)不動(dòng)點(diǎn),故只能有①f(t)t,這與f(x)只有a,b兩個(gè)不動(dòng)點(diǎn).②f(t)a或b不妨設(shè)f(t)a,則f(f(t))f(a)at,7、試求出一個(gè)整系數(shù)多項(xiàng)式f(x)axn xn1...axa，使得方程f(x)0 23一根 2323解:令x ,得(x23

2)33x36x3

2(3x2

兩邊平方,x612x46x336x236x918x424x2f(x)x66x46x312x236x1即可8x43x22x3表示成兩個(gè)不同次數(shù)的整系數(shù)多項(xiàng)式的平方差，并求出所解:f(xx2axbg(xcxd(cf2g2x2axb)2cxd)2x43x22x3x42ax3(a22bc2)x2(2ab2cd)x(b2d2)2a有方程組a22bc2有方程組2ab2cd

a解得b解得c

,f(xx22g(xx另三組解為(fg),(fg),(f9、(2006交大)設(shè)k9xx32kx2k2x9k27解:看作關(guān)于k的二次方程

xk2(2x29)k(x327)0

(6x9)22(x29)(6x于是k 得

k3)x90xk3,x(3k)

(k9)(k3)(k9)xk3210（34頁）解方程(x216)(x3)29x23x 解:x3不是根,x2 x3

3x x2 xx3

166 x

x3

166 x得x

進(jìn)而x422i或x 11a1a2,...a100b1b2,...,b100為互不相同的實(shí)數(shù)，將它們按照如下法則填入100100,100方格表：第ij列的方格填入aibj.1，求證：任何一行數(shù)的乘積都等于1,100證:構(gòu)造多項(xiàng)式f(x)(xa1)(xa2 (xa100)

由題

f(bi)0

(xa100)1(xb1)(xb2 又f(x)是首項(xiàng)系數(shù)為1的多項(xiàng)式,且degf(x)100.f(x)(xb(xa100)1(xb1)(xb2 令xai,i1, ,100即證

12、{a1a2an},{b1b2bn}是的兩個(gè)不同的n{aiaj|1i

jn}{bibj|1i

jnr{aiaj|1i

jn}滿足aia

r(1i

jn的數(shù)組(i,j的個(gè)數(shù)和滿足bibjr(1i

jn(i,j的個(gè)數(shù)相等.n2的方冪證:ai,bj

,1ij

事實(shí)上,將2n個(gè)數(shù)乘以它們的公分母,代替原來個(gè)數(shù),仍滿足題設(shè)條件;同時(shí)可不妨設(shè)它們都是正整數(shù),因它們每個(gè)數(shù)都減去最小數(shù)再加仍滿足題設(shè)條件.于是設(shè)aibj

*(1in,1jn).定 f(x)xai,g(x)xbj

(f(x))2(g(x))2x2ai

f(x2)g(x2 f(xg(xx1)kh(xk0,h(x為整系數(shù)多項(xiàng)式,h(10f(x2)g(x2)(f(x)g(x))(f(x)(x21)kh(x2(f(xg(x))(x1)kh(x對(duì)x成立(x1)kh(x2)(f(x)

x

得n2k1得證(2003復(fù)旦)x

log axa或(2006交大)已知abcbca(bc)x2b(ca)xc(ab)0a

,1b

1成等差數(shù)列c證:可將配成(abbc2ac)2

f(

)f(1設(shè)f(x)rr

,如果

1003，求1

的值

12

frrr 解:f(xax3ax2ax

3則 33rrr312 以及f1a3a2a1a,f1a3a2a1a2

2

2 1a于是1003

0

從而原式r1r2r3a22002 P使得三次方程5x35(P1)x271P1)x166P的三個(gè)根均為自解:x1是一個(gè)根,故降次為二次方程5x25Px66P1uv設(shè)兩根為u,v(uv),則

,①代入②得5uv66(uv1uv5(66P知uv23,11整除,故由上式

v66u1,因uv

,得u66u52u

u,知u

vu可得5u2132u1

這樣17u

u17,19,23,25,逐一檢驗(yàn),僅u17v59.(P76)滿足題意f(x)

x2007

x2006...ax32x2x1

證:f(0)1f(x的根不為零.y1,x g(y)y2007y20062y2005ay2004 y 設(shè)y,y , 是g(y)的根,則由韋達(dá)定 y2

2 2

yy1 1y于是必有某個(gè)yi為虛數(shù),從而xi 是虛數(shù)1yi

i

i1i12實(shí)數(shù)a,bc和正數(shù)使得f(x)x3ax2bxc有三個(gè)實(shí)根x12

,

滿足x2x1;

x1(xx.

2a327c

1解:x32(x1x2t(t0,由韋達(dá)定理12a327c 2(xxx)327xxx9(xxx)(xxxxxx 12 1 2 3 (x1x22x3)(x2x32x1)(x3x12x2t(2t3)(3 不妨設(shè)9

1414(924t2)(924t2)8t 2(924t2)8t243 2t 3”取到.此時(shí)a,bc都可用三根表示,不羅列2.不等式的證明,不僅需要應(yīng)用不等式的基本性質(zhì)、定理和已證的不等式,有時(shí)還需要靈活交替地運(yùn)用一些證明不等式的方法.最常用的方法有:比較法;綜合法;分析法;放縮法;反證法等..證明

x8x5x2x10,x證:LHSx41x)23x2x (x41x)23(x2)22 xy.x2y2xy3(xy證:LHSRHSx2y2xy3x3yx2(y3)xy23y(xy3)23y23y (xy3)23(y1)2*,證明*,證明已知n

1

1

3nn(n n 2(nn(n n 2(n nn

22(n n2(n nnn n LHS

2 11212131n1212131nnn3n

證明:對(duì)任意正實(shí)數(shù)abc,有

1a3b3 b3c3 c3a3 證:a3b3ab)(a2b2aba a3b3 ab(ab) abc(ab同理

b3c3 abc(ab上述三式相加,即得結(jié)論

c3a3 abc(ab求證

k116k1kkkkk

證:注意到

k)1

k1)

kk11)kk1k

k1k1k

1

又1

1)1

1)

故(*)得證設(shè)abcd0.S

證明:1Sab bc cd da證:S

abc abc abc abcS

故1Sa a c c設(shè)abc

證明

bc

ca

ab

(1a)(1b)(1c)證:不妨設(shè)abcLHS

ab

ab

ab

(1a)(1b)(1abc(ab1)(1a)(1b)(1ab ab1abc(1a)(1a)(1b)(1b)(1c)abc

1ab1

ab

ab

ab1已知ab

,a3b3

證明

ab證:假設(shè)ab

a2

即a32故a3b3(2b)3b3812b6b26(b1)222,設(shè)abc

abc

求證

a2b2c2證:a2b2c2

則a2

a同理b

c

因此

abcabbca2b2

b2c2

c2a2于是a2b2c2abbc故abca2b2c2abbccaabc. 證明已知xyz 證明

:下列四個(gè)不等式中,至少有一個(gè)不等式不成立|x||yzt

|y||xzt

|z||xyt

|t||xyz|證:假設(shè)四個(gè)不等式都成立,x2(yzy2(xzz2(xy

(xyzt)(xyzt)(yxzt)(yxzt)(zxyt)(zxyt) t2(xyabc

a2b2c21, 2(a3b3S 的最小值

解:abc時(shí),S3.下證S3 2(a3b3S3 3 11

2(a3b3c3

a2b2c2a2b2c2a2b2c2a2b2c2

2 ab2 1

2 1

2 1 c2ab2c2

c2a2

a2b22 ab 1 1

1a2 b2 c2 c a bSminABC為ABC的三個(gè)內(nèi)角,求證cosBcosC

b

A4sin A

2cosB證:(*2cosBCcosBC

cosB2

4cosB22cos2BC4cosBC2

B2

xyz0xyz

求證xyx2y2z2xyyzzx zxy證:LHSx2y2z211 x21y21z2 xyz xyzSmax{a21,b1c2,S的最小值 解:4S2a21b1c2 c a 2a2b4c12 a1,b2c1時(shí),”成立設(shè)abc0且5a44b46c4

S求5a32b33c3的最大值解:a31(aaa21(a4a4a4243a4 同理b33b41

c33c41 故5a32b33c33(5a44b46c445 a2,bc1時(shí),”成立習(xí)題xyzxyyzzx

求證

x25y28z2證明

1352n12n 2n1

2n1,證明設(shè)x,y 5x26xy9y212x12y8,證明求證:xyz,下述三個(gè)不等式不可能同時(shí)成立:|x||yz|,|y||zx|z||xy|設(shè)abc0,證明:xyz,有x2y2z2

abxy bcyz cazx(ab)(bc(ab)(bc)(c 已知ab

證明1

1 1b

,p,gk1(x)g(gk(x))若gk(xTx(k,p,

pg(x)的迭代周期.

g(xyx對(duì)稱,g(g(x))g(g1(x))

y2xx

yxx

y12

f為定義在區(qū)間I上的函數(shù),若對(duì)I上任意兩點(diǎn)x1,x2和 總f(x11)x2f(x11f(x2.fI上的下凸函數(shù)特別地

1時(shí),fx1x2f(x1f(x2.”,則稱上凸函數(shù) fI上二階可導(dǎo),f是凸函數(shù)下凸函數(shù)的n3情形

f''(x)nf在[ab．,則對(duì)xi[a

i0(i1 n且iifxiii

if(xi ABC中ABC中

sinAsinBsinC32ysinx在(0,上下凸,故sinAsinBsinCsinABC e.g.ylnx在(0．(y0),lna1lna1lna2 lnanlna1a2 anna aa1a2 an此即均值不等式1 f(x的條件(f(x的類型f(x解析式,或是不求解析式,求一些其他的結(jié)論f(sinx1)cos2x2換元.ysinx1[20],fyy22y2y[2證:即證當(dāng)0時(shí),有tan1(tantan sin(*)

1(cossincossin)1sin() 2

coscos

2coscoscoscoscos21(1cos( coscos1(1coscossinsin2cos()1.這顯然成立

(11xyf(xy)f(x)f(解首先f(0)于是f(nn(n

(令xy0f(n1f(n1(令xny)另令xnynf(nn(n)對(duì)x

x

f(nf(n)f(n)mmf(n)=f(mmm

且(|m|,|n|1

m0mf(x)

)f(n)于是f(xx(x

2xyf(xy)f(x)f(y)f(xy)解y1f(xf(xf(1f(x11f(12,f(x1f(x,f(n,f(n)1

x

,yn,(nn

n0f

n)f()f(n)f(n

) 得f()

再取xpy1pq,q,qfp1fpf1fp11p111p1p f(xx1.驗(yàn)證后知滿足已知f(x1xn2

f1(x)f

(x)f(

(x))，如果f13(x)f31(x)，那么f16(x)的解析式 x注意到f(x)的迭代周期是6.可得f16(x)f4(x) x1x，已知函數(shù)f(x)1x，

fn(x)

fff(x)))(n個(gè)f

f99

解:f(1x)

f(x)

,f(2)(x)f(f(x)) 11,111,1

f(99)(1)1.得Mf(x的全體：存在非零常數(shù)Tx得f(xT)Tf(xf(xx是否屬于集合Mf(x)ax(a0a1yxf屬于f(xsinkxM,求k的取值范圍,Tf解:(1)設(shè)f(x)M,則T0,對(duì)x f(x,TfxTTx,x0知T不存在,f(xMf(xax,f(xTaxT.若axTTax,則(aTT)ax0對(duì)xaTT0f(xaxyx圖像有公共點(diǎn)T使aT0T 取TT0)由題意知f(xT)Tf(x),即sink(xT)Tsinkx對(duì)x ．:左右兩邊函數(shù)值域相等,于是T1.(k0時(shí))進(jìn)而可得kn(n,n求解函數(shù)方程f(nmf(nmf(3n,n解:令m0得2f(n)f(3n),n ,令mn0得f(0)mnf(2nf(0f

f(2nf

進(jìn)而對(duì)m,f(4m)f(6m)f,于是f(2m0(m

另一方面,令n3mf(4mf(2mff(n1f(3n1f(2n0f(n0 求所有函數(shù)f ，使得對(duì)任意實(shí)數(shù)x,y,z，1f(xy)1f(xz)f(x)f(yz) 解:xyz1f(1f(1))214

(f(1)1)22f(11.yz1f(x1f(x1 f(x12xyz0f(0f2(01f(01x011fyz1

f(yz)12

f(x12f(x12f

xy(xy)f(xy)(xy)f(xy)4xy(x2y2解:(xyf(xyxyf(xy)4xy(x2y2xy0f(0設(shè)uxyvxy,則uv2x,uv2①式變?yōu)関f(uuf(vuv(u2v2.當(dāng)uv0f(u)

f(v)u2

f(u)u2

f(v)

v2= f(x)x3Cx(x

f(0)0x0f(xx3Cx(C是常數(shù)

f

xy

f(xyf(xfy4xyf(-1)f(14f解:xy0.f(xyf(xfy4xy①f(0)0x1y1,f(1f(14.f(-1)f(14,f(1)0,f(1)0,又由4f(1)f(1) 4,f(1)f(1)y1,f(x1)f(x4x(n1))2(n1)2n2.n(n1))2(n1)2n2.n*又fk1fkf14k fk1fkf14k n n n n n n

k從1n1f12n nm 同理對(duì)k從1m1fn2n2 m m m 最后令x ,y ,由

f(0)f f f 2 ,f,f(x) m 2 n

綜上,對(duì)

2x f

f

x2y2)

f(x)f(y)(x,y

f(x)(f(1))