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1、不 等 式 练 习 题第一部分1下列不等式中成立的是（）A若 ab，则22acbc B若 ab，则22abC若0ab，则22aabb D若0ab，则11ab2已知113344333,552abc，则,a b c的大小关系是（）(A).cab (B)abc (C)bac (D)cba3已知,a b c满足 cba且0ac，下列选项中不一定成立的是（）（A）abac（B）0c ba（C）22cbab（D）()0ac ac4规定记号“”表示一种运算，定义ab=baab（a,b为正实数），若 1k2b，则1abc2，则 ab；若 a|b|，则 ab；若 ab，则 a2b2.其中正确的是()AB CD
2、2设 a，bR，若 a|b|0，则下列不等式中正确的是()Aba0 Ba3b20 Da2b20)Cysin xcscx，x(0，2)Dy7x7x4已知 loga(a21)loga2a0，则 a 的取值范围是()A(0,1)B(12，1)C(0，12)D(1，)5f(x)ax2ax1 在 R上满足 f(x)0，b0.若3是 3a与 3b的等比中项，则1a1b的最小值为()A8 B4 C1 D.148 已知当 x0时，不等式 x2mx 40恒成立，则实数 m的取值范围是 _9.已知 Ax|x23x20，B x|x2(a1)xa0(1)若 AB，求 a的取值范围；(2)若B?A，求a的取值范围10.
3、已知 x0，y0，且1x9y1，求 xy 的最小值11已知 a，b，c 都是正数，且 abc1.求证：(1 a)(1 b)(1 c)8 abc.证明a、b、c 都是正数，且 abc1，1abc2 bc0，1bac2 ac0，1cab2 ab0.(1a)(1 b)(1 c)2 bc2 ac2 ab8abc.12不等式 kx22x6k0(k0)(1)若不等式的解集为 x|x2，求 k 的值；(2)若不等式的解集为R，求 k 的取值范围文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4
4、A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7
5、 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L
6、10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P1
7、0K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW
8、2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F
9、4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4
10、M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4参 考 答 案第一部分1D【解析】对于 A，若0c，显然22acbc不成立；对于 B，若0ba，则22ab不成立；对于 C，若0ab，则22a
11、abb，所以 C错；对于 D，若0ab，则10ab，所以11ab；故选 D 2D【解 析】因 为11034所 以110343331555即1ab，且30433122所以1c，综上，cba，所以答案为：D.3C【解析】,0,0,0ac acca.(1),0bc a,abac;(2),0baba,0,0cc ba;(3),0caac,0,0acac ac.(4)cba 且0,0ca,0b或0b或0b,2cb和2ab的大小不能确定,即C选项不一定成立.故选 C.4A【解析】根据题意2221113kkk化简为220kk，对 k 分情况去绝对值如下：当0k时，原不等式为220kk解得21k，所以 01k
12、；当0k时，原不等式为20成立，所以0k；当0k时，原不等式为220kk，解得12k，所以10k；综上，11k，所以选择 A.5B【解析】对于 A，当0c时，不等式不成立，故A错；对于 C，因为0ab，两边同时除以0ab，所以11ab，故 C错；对于 D，因为0ab，110ba，所以abba，故 D错，所以选 B文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K
13、10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编
14、码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4
15、A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7
16、 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L
17、10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P1
18、0K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW
19、2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F46A【解析】0.53422，ablogclog，0.5122112，341122，loglog bac 故选：A7C【解析】根据题意化简不等式为()(1()1xaxa,即22(1)0 xxaa对任意实数x成立,所以根据二次恒成立0,解得2321a81【解析】由2
20、4xy化为42xy代入14yxy得4111111212 428248xxyxyxyxy151428yxxy，因为0,0 xy，所以1151151214428428yyxy xxyxyx y（当且仅当“43xy”时，取“”），故最小值为 1.9xyacyxyxcyxyxa222222222,；0,0,0 xyyx，即ac；10（1）2a（2）132xx【解析】（1）由2M，说明元素 2 满足不等式2520axx，代入即可求出a的取值范围；（2）由122Mxx，1,22是方程2520axx的两个根，由韦达定理即可求出2a，代入原不等式解一元二次不等式即可；（1）2M，225 220a，2a（2）1
21、22Mxx，1,22是方程2520axx的两个根，文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K
22、10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编
23、码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4
24、A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7
25、 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L
26、10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P1
27、0K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW
28、2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4由韦达定理得15221222aa解得2a不等式22510axxa即为：22530 xx其解集为132xx第二部分2.解析由 a|b|0?|b|a?ab0，故选 C.3.解析yx22x的值域为(，2 2，)；yx2x1x11x12(x0)；ysin xcscxsin x1sin x2(0sin x0，b0，abab212?ab14.1a1babab1ab1144.11.解析因为x0，y0，1x9y1，所以 xy(xy)(1x9y)yx9xy102yx9xy1016.当且仅当yx9xy时，等号成立，又因为1x9y1.所以当
29、x4，y12 时，(xy)min16.文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K
30、4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R
31、5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文
32、档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6
33、X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3
34、Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL
35、5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4文档编码:CW2R5K10P10K4 HL5L10D3J3Q7 ZF4M6X4A10F4