新版高中數(shù)學(xué)北師大版必修5習(xí)題第三章不等式3.3.2.1

3.2基本不等式與最大(小)值第1課時(shí)利用基本不等式求最值課時(shí)過關(guān)·能力提升1.設(shè)a>0,b>0,若3是3a與3b的等比中項(xiàng),則1a+1bA.8 B.4 C.1 D.1解析:因?yàn)?a·3b=3,所以a+b=1,所以1a+1b=(a+b)1a+1b=2+ba+ab≥2+2b答案:B2.若x>0,則y=33x1x的最大值是(A.3 B.332 C.323 D.1解析:y=33x1x=33x+1x≤323x·1x=323,當(dāng)且僅當(dāng)3x=答案:C3.已知a>0,b>0,則1a+1b+2abA.2 B.22 C.4 D.5解析:1a+1b+2ab≥21ab+2ab≥當(dāng)且僅當(dāng)1a=1b,1ab=ab,即a=b=1時(shí),答案:C4.若x>1,則函數(shù)y=x2+11-x2A.3 B.0 C.1 D.2解析:當(dāng)x>1時(shí),x2+1x2-1=x21+1x2-1+1≥2(x2-1)·1x2-1答案:A5.若a>0,b>0,a+b=2,則y=1a+4bA.72 B.4 C.92 D解析:∵a+b=2,∴1=a+b2,4=2(∴y=1a+4b=a+b2a+2(a+b)b答案:C6.若p>0,q>0,p,q的等差中項(xiàng)是12,x=p+1p,y=q+1q,則x+y的最小值為A.6 B.5 C.4 D.3解析:由題意知p+q=1,p>0,q>0,∴x+y=p+q+1p+1q=1+1pq≥1當(dāng)且僅當(dāng)p=q=12時(shí),等號(hào)成立答案:B7.若2x+2y=1,則x+y的取值范圍是()A.[0,2] B.[2,0]C.[2,+∞) D.(∞,2]解析:∵2x+2y≥22x·2y∴2x+y≤1∴x+y≤2.故選D.答案:D8.已知函數(shù)f(x)=4x+ax(x>0,a>0)在x=3時(shí)取得最小值,則a=.解析:∵4x+ax≥24a,當(dāng)且僅當(dāng)4x=ax時(shí),等號(hào)成立,∴4x2=a,∴a=4×32答案:369.若a+b=2,b>0,則12|a|解析:12|a|+|a|b=a+b4|a|+|a|b=故12|a答案:310.若實(shí)數(shù)x,y滿足x2+y2+xy=1,則x+y的最大值是.

解析:∵x2+y2+xy=1,∴(x+y)2=xy+1.又xy≤x+∴(x+y)2≤x+y22+1,即34(x+y∴(x+y)2≤43,當(dāng)且僅當(dāng)|x|=|y|=33時(shí),∴233≤x+y≤∴x+y的最大值為23答案:2★11.求下列函數(shù)的最值.(1)y=x(25x),x∈0,(2)y=x·3-x2,x∈(0,解:(1)∵x∈0,∴x>0,25x>0,y=x(25x)=15·5x(25x≤15當(dāng)且僅當(dāng)5x=25x,即x=15時(shí),等號(hào)成立∴y=x(25x)的最大值為15(2)∵x∈(0,3),∴x>0,3x2>0.∴y=x·3-當(dāng)且僅當(dāng)x=3-即x=62時(shí),等號(hào)成立∴y=x·3-x2★12.當(dāng)x>12時(shí),求函數(shù)y=x+82x-1的最小值解:y=x+82