新版高中數(shù)學(xué)人教A版必修5習(xí)題第三章不等式3.4.1

3.4基本不等式:第1課時基本不等式課時過關(guān)·能力提升基礎(chǔ)鞏固1若x>0,則x+4A.2 B.3 C.2答案:D2若x,y滿足x+y=40,且x,y都是正數(shù),則xy的最大值是().A.400 B.100 C.40 D.20解析:xy≤x+y22=400,答案:A3若0<x<1A.解析:∵0<x<∴y=x(13x)=≤1當(dāng)且僅當(dāng)3x=13x,即x=1答案:B4設(shè)a,b∈R,若a≠b,a+b=2,則必有().A.1≤ab≤aC.ab<解析:令a=1,b=3,則ab=3,a2+所以排除選項A,C,D,故選B.答案:B5若M=a2+4a(a∈R,a≠0),則MA.(∞,4]∪[4,+∞) B.(∞,4]C.[4,+∞) D.[4,4]解析:當(dāng)a>0時,M=當(dāng)且僅當(dāng)a=4a,即a=2時取當(dāng)a<0時,M==a+≤2當(dāng)且僅當(dāng)a=-4a,即a=2時取綜上,M的取值范圍為(∞,4]∪[4,+∞).答案:A6若a>b>1,P=lgA.R<P<Q B.P<Q<RC.Q<P<R D.P<R<Q解析:∵a>b>1,∴l(xiāng)ga>0,lgb>0.∴R=lgaQ>答案:B7若a>0,b>0,則2解析:2即a=2b時取答案:28當(dāng)函數(shù)y=x2(2x2)取最大值時,x=.

解析:當(dāng)-2<x<2時,≤x當(dāng)且僅當(dāng)x2=2x2,即x=±1時,等號成立,當(dāng)x2≥2時,y=x2(2x2)≤0,不可能取最大值.所以當(dāng)x=±1時,y=x2(2x2)有最大值為1.答案:±19已知2解∵x>0,y>0,∴2=2x+3y≥26xy(當(dāng)x=2,∴xy≥6,從而xy≥6,10已知x>1,試求函數(shù)y=x解∵x>1,∴x+1>0,∴y==x+1+當(dāng)且僅當(dāng)x+1=4x+1,即x=所以函數(shù)y=x2能力提升1若2a+b=1,a>0,b>0,則1A.2解析:1=2+1+∵a>0,b>0,∴≥3+2當(dāng)且僅當(dāng)ba=2ab∴1a+答案:C2若x+3y2=0,則函數(shù)z=3x+27y+3的最小值是().A.3解析:z=3x+27y+3≥2∵x+3y2=0,∴x+3y=2.∴z≥23x+3y+3=232+3=9,當(dāng)且僅當(dāng)3答案:D3若a>0,b>0,a+b=2,則y=1A.解析:依題意得1a+4答案:C4當(dāng)x>1A.解析:∵x>∴y=x+≥2當(dāng)且僅當(dāng)x-12=答案:A5設(shè)a,b>0,a+b=5,則a解析:因為a,b>0,a+b=5,所以(a+1)+(b+3)=9.令x=a+1,y=b+3,則x+y=9(x>1,y>3),于是a+1+b+3=x+y,而(x+y)2=x+y+2xy≤x+y+(x+y)=18,所以x+y≤32.此時x=y,答案:3★6函數(shù)y=loga(x1)+1(a>0,且a≠1)的圖象恒過定點A,若點A在一次函數(shù)y=mx+n的圖象上,其中m,n>0,則1解析:由題意,得點A(2,1),則1=2m+n.又m,n>0,所以=4+nm+4當(dāng)且僅當(dāng)nm=則1m答案:8★7若對任意x>0,xx2+3x+1≤a恒成立,則解析:因為x>0,所以x+1x≥當(dāng)且僅當(dāng)x=1時取等號,所以有x即xx2答案:1★8已知f(x)=ax(a>0,且a≠1),當(dāng)x1≠x2時,比較fx解∵f(x)=ax,∴f∴∵a>0,且a≠1,x1≠x2,∴∴即9若正實數(shù)x,y滿足2x+y+6=xy,求xy與2x+y的最小值.解∵2x+y+6=xy,x>0,y>0,∴xy=2x+y+6≥2即xy22xy-當(dāng)且僅當(dāng)2x=∴(xy-3∵xy+2又2x+y+6=∴(2x+