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2017-2018学年高中数学北师大版必修5课时作业:第三章 不等式 18

课时作业 18 基本不等式 |基础巩固|(25 分钟,60 分) 一、选择题(每小题 5 分,共 25 分) 1.不等式(x-2y)+x-12y≥2 成立的条件为( ) A.x≥2y,x-2y=1 B.x>2y,x-2y=1 C.x≤2y,x-2y=1 D.x<2y,x-2y=1 解析:因为不等式成立的前提条件是各项均为正,所以 x-2y>0,即 x>2y,且等号成立 时(x-2y)2=1,即 x-2y=1,故选 B. 答案:B 2.已知 m=a+a-1 2(a>2),n=2 2-b2 (b≠0),则 m,n 之间的大小关系是( ) A.m>n B.m<n C.m=n D.不确定 解析:因为 a>2,所以 a-2>0, 又因为 m=a+a-1 2=(a-2)+a-1 2+2, 所以 m≥2 a- 1 a-2+2=4, 由 b≠0,得 b2≠0,所以 2-b2<2,n=2 2-b2 <4. 所以 m>n. 答案:A 3.若 a+b=1,恒有( ) A.ab≤14 B.ab≥14 C.a2b2≤16 D.以上均不正确 解析:因为 a+b=1>0, 所以 a,b 中至少有一个为正数. 故当 a,b 中有一个是负数或 0 时,显然有 ab≤0<14; 当 a,b 均为正数时,有 1=a+b≥2 ab, 所以 ab≤14. 答案:A 4.已知 a,b∈(0,1),且 a≠b,下列各式中最大的是( ) A.a2+b2 B.2 ab C.2ab D.a+b 解析:因为 a,b∈(0,1),所以 a2<a,b2<b, 所以 a2+b2<a+b,又 a2+b2>2ab(因为 a≠b), 所以 2ab<a2+b2<a+b. 又因为 a+b>2 ab(因为 a≠b),所以 a+b 最大.故选 D. 答案:D 5.设 a,b 为正数,且 a+b≤4,则下列各式中正确的一个是( ) A.1a+1b<1 B.1a+1b≥1 11 11 C.a+b<2 D.a+b≥2 解析:因为 ab≤???a+2 b???2≤???42???2=4,所以1a+1b≥2 a1b≥2 14=1. 答案:B 二、填空题(每小题 5 分,共 15 分) 6.设 a>1,且 m=loga(a2+1),n=loga(a+1),p=loga(2a),则 m,n,p 的大小关系 是________(用“>”连接). 解析:因为 a>1,所以 a2+1>2a>a+1, 所以 loga(a2+1)>loga(2a)>loga(a+1), 所以 m>p>n. 答案:m>p>n 7.设正数 a,使 a2+a-2>0 成立,若 t>0,则12logat________logat+2 1(填“>”“≥”“≤” 或“<”). 解析:因为 a2+a-2>0,所以 a<-2 或 a>1, 又 a>0,所以 a>1, 因为 t>0,所以t+2 1≥ t, t+1 所以 loga 2 ≥loga t=12logat. 答案:≤ 8.给出下列不等式: ①x+1x≥2;②???x+1x???≥2;③x2+ xyy2≥2; ④x2+2 y2>xy;⑤|x+2 y|≥ |xy|. 其中正确的是________(写出序号即可). 解析:当 x>0 时,x+1x≥2;当 x<0 时,x+1x≤-2,①不正确; 因为 x 与1x同号,所以???x+1x???=|x|+|1x|≥2,②正确; 当 x,y 异号时,③不正确; 当 x=y 时,x2+2 y2=xy,④不正确; 当 x=1,y=-1 时,⑤不正确. 答案:② 三、解答题(每小题 10 分,共 20 分) 9.设 a,b,c 为正实数,求证:(a+b+c)·???a+1 b+1c???≥4. 证明:因为 a,b,c 为正实数, 所以(a+b+c)·???a+1 b+1c??? =[(a+b)+c]???a+1 b+1c??? =1+a+c b+a+c b+1≥2+2 a+c b·a+c b=2+2=4. c a+b 当且仅当a+b= c ,即 a+b=c 时,取等号. 所以(a+b+c)·???a+1 b+1c???≥4. 10.设 a,b,c 均为正数,且 a+b+c=1.证明: (1)ab+bc+ac≤13; a2 b2 c2 (2) b + c + a ≥1. 证明:(1)由 a2+b2≥2ab,b2+c2≥2bc,c2+a2≥2ca, 得 a2+b2+c2≥ab+bc+ca. 由题设得(a+b+c)2=1, 即 a2+b2+c2+2ab+2bc+2ca=1. 所以 3(ab+bc+ca)≤1,即 ab+bc+ca≤13. (2)因为ab2+b≥2a,bc2+c≥2b,ca2+a≥2c, a2 b2 c2 故 b + c + a +(a+b+c)≥2(a+b+c), 即ab2+bc2+ca2≥a+b+c. a2 b2 c2 所以 b + c + a ≥1. |能力提升|(20 分钟,40 分) 11.若 f(x)=???12???x,a,b 均为正数,P=f???a+2 b???,G=f( ab),H=f???a2+abb???,则( ) A.P≤G≤H B.P≤H≤G C.G≤H≤P D.H≤G≤P 解析:因为 a,b 均为正数, 所以a+2 b≥ ab= aabb≥aa+bb=a2+abb,又因为 f(x)=???12???x 为减函数,所以 f???a+2 b??? 2 ≤f( ab)≤f???a2+abb???,所以 P≤G≤H. 答案:A 12.若 a>0,b>0,a+b=2,则下列不等式①ab≤1;② a+ b≤ 2;③a2+b2≥2; ④1a+1b≥2,对满足条件的 a,b 恒成立的是________.(填序号) 解析:因为 ab≤???a+2 b???2=1,所以①

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