专题强化练1 利用基本不等式求最值-2020-2021学年高一数学同步练习和分类专题教案（人教A版2019必修第一册）

第二章一元二次函数、方程和不等式专题强化练1利用基本不等式求最值一、选择题1.已知实数a>0,b>0,1a+1+1bA.32 B.22 C.3 D.22.若正数a,b满足1a+1b=1,则1a-1A.3 B.4 C.5 D.63.设a>b>0,则a2+1ab+1aA.1 B.2C.3 D.44.设x,y都是正数,且xy-(x+y)=1,则()A.x+y≥2(2+1) B.xy≤2+1C.x+y≤(2+1)2 D.xy≥2(2+1)5.若正数x,y满足x+y+15=1x+9yA.x为定值,但y的值不确定B.x不为定值,但y是定值C.x,y均为定值D.x,y的值均不确定6.已知x>0,y>0,2x-1x=8yA.2 B.22 C.32 D.47.若正数x,y满足x+3y=5xy,当3x+4y取得最小值时,x+2y的值为()A.245 C.285 二、填空题8.已知正数x,y满足x+2y=2,则x+8yxy9.若实数a>1,b>2,且满足2a+b-6=0,则1a-1+2b10.若a,b∈R,且a2+2ab-3b2=1,则a2+b2的最小值为.

三、解答题11.已知正数a,b满足a+b=4,求1a+1+12.(1)设a>b>c,且1a-b+1(2)记F=x+y-a(x+22xy答案全解全析一、选择题1.B∵a>0,b>0,∴a+1>1,b+1>1.又∵1a+1+1b+1=1,∴a+2b=[(a+1)+2(b+1)]-3=[(a+1)+2(b+1)]·1a+1+1b+1-3=1+2(b+1)a+1+2.B∵a>0,b>0,1a+1b∴a>1,b>1,a+b=ab,∴1a-1>0,∴1a-1+4=24ab当且仅当1a-1=4b3.D∵a>b>0,∴a-b>0,∴a2+1ab+1a(a-b)=a2-ab+ab+1ab+1a(a-b)=a(a-b)+4.A∵x>0,y>0,且xy-(x+y)=1,∴xy=1+(x+y)≥1+2xy(当且仅当x=y=1+2时,等号成立),即(xy)2-2xy-1≥0,解得xy≥1+2,即xy≥(1+2)2.xy=1+(x+y)≤(x+y即(x+y)2-4(x+y)-4≥0,解得x+y≥2(2+1).故选A.5.C由题得(x+y)1x+9y=1+9+yx+9xy≥1+9+2yx·9xy=16,因为x+y≤1,所以x+y+15=1x+96.C2x-1x=8y-y⇒2x+y=1x要求2x+y的最小值可以先求(2x+y)2的最小值.(2x+y)2=(2x+y)(2x+y)=(2x+y)·1x+8y=2+16xy+yx+8=10+16xy+yx≥216xy·7.B∵x+3y=5xy,x>0,y>0,∴15y+∴3x+4y=(3x+4y)15y+35x=135+3当且仅当3x5y二、填空题8.答案9解析因为x,y为正数,且x+2y=2,所以x2+y=1,所以x+8yxy=1y+8x·x2+y=9.答案4解析∵a>1,b>2,且满足2a+b-6=0,∴2(a-1)+b-2=2,a-1>0,b-2>0,则1a-1+2b-2=1≥1=12当且仅当b-2a-1=4(a-1)b-2故答案为4.10.答案5解析由a2+2ab-3b2=1得(a+3b)(a-b)=1,令x=a+3b,y=a-b,则xy=1且a=x+3y4所以a2+b2=x+3y=x2+5y2+2xy当且仅当x2=5,y2=55故答案为5+1三、解答题11.解析∵a+b=4,∴(a+1)+(b+3)=8,∴81a+1+1b+3=[(a+1)+(b+3)]·1a∴1a+1+1b∴1a+1+1b12.解析(1)由a>b>c,知a-b>0,a-c>0,所以原不等式等价于a-ca要使原不等式恒成立,只需a-ca因为a-ca