專題強(qiáng)化練1 利用基本不等式求最值-2020-2021學(xué)年高一數(shù)學(xué)同步練習(xí)和分類專題教案（人教A版2019必修第一冊(cè)）

第二章一元二次函數(shù)、方程和不等式專題強(qiáng)化練1利用基本不等式求最值一、選擇題1.已知實(shí)數(shù)a>0,b>0,1a+1+1bA.32 B.22 C.3 D.22.若正數(shù)a,b滿足1a+1b=1,則1a-1A.3 B.4 C.5 D.63.設(shè)a>b>0,則a2+1ab+1aA.1 B.2C.3 D.44.設(shè)x,y都是正數(shù),且xy-(x+y)=1,則()A.x+y≥2(2+1) B.xy≤2+1C.x+y≤(2+1)2 D.xy≥2(2+1)5.若正數(shù)x,y滿足x+y+15=1x+9yA.x為定值,但y的值不確定B.x不為定值,但y是定值C.x,y均為定值D.x,y的值均不確定6.已知x>0,y>0,2x-1x=8yA.2 B.22 C.32 D.47.若正數(shù)x,y滿足x+3y=5xy,當(dāng)3x+4y取得最小值時(shí),x+2y的值為()A.245 C.285 二、填空題8.已知正數(shù)x,y滿足x+2y=2,則x+8yxy9.若實(shí)數(shù)a>1,b>2,且滿足2a+b-6=0,則1a-1+2b10.若a,b∈R,且a2+2ab-3b2=1,則a2+b2的最小值為.

三、解答題11.已知正數(shù)a,b滿足a+b=4,求1a+1+12.(1)設(shè)a>b>c,且1a-b+1(2)記F=x+y-a(x+22xy答案全解全析一、選擇題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當(dāng)且僅當(dāng)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(當(dāng)且僅當(dāng)x=y=1+2時(shí),等號(hào)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).故選A.5.C由題得(x+y)1x+9y=1+9+yx+9xy≥1+9+2yx·9xy=16,因?yàn)閤+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當(dāng)且僅當(dāng)3x5y二、填空題8.答案9解析因?yàn)閤,y為正數(shù),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,且滿足2a+b-6=0,∴2(a-1)+b-2=2,a-1>0,b-2>0,則1a-1+2b-2=1≥1=12當(dāng)且僅當(dāng)b-2a-1=4(a-1)b-2故答案為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,則xy=1且a=x+3y4所以a2+b2=x+3y=x2+5y2+2xy當(dāng)且僅當(dāng)x2=5,y2=55故答案為5+1三、解答題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,所以原不等式等價(jià)于a-ca要使原不等式恒成立,只需a-ca因?yàn)閍-ca