2018-2019版高中數(shù)學(xué) 第三章 不等式 3.4.1 基本不等式練習(xí) 新人教A版必修5.doc

第1課時(shí)基本不等式課后篇鞏固探究A組1.若a0,b0,且a+b=2,則下列不等式正確的是()A.ab1B.ab1C.a2+b24D.a2+b24解析由已知可得aba+b22=1,而a2+b2=(a+b)2-2ab=4-2ab2,故只有A正確.答案A2.若x>0,y>0,且x+y=,則xy的最大值為()A.233B.23C.D.136解析由基本不等式可得xyx+y22=1322=136,當(dāng)且僅當(dāng)x=y=時(shí),xy取最大值136.答案D3.若實(shí)數(shù)a,b滿足a+b=2,則3a+3b的最小值是()A.18B.6C.23D.243解析3a+3b23a3b=23a+b=232=6,當(dāng)且僅當(dāng)a=b=1時(shí),取等號(hào).故3a+3b的最小值是6.答案B4.已知a,b均為正實(shí)數(shù),則下列不等式不一定成立的是()A.a+b+1ab22B.(a+b)1a+1b4C.a2+b2aba+bD.2aba+bab解析A項(xiàng),a+b+1ab2ab+1ab22,當(dāng)且僅當(dāng)a=b=22時(shí)等號(hào)同時(shí)成立;B項(xiàng),(a+b)1a+1b=2+ab+ba4,當(dāng)且僅當(dāng)a=b時(shí)取等號(hào);C項(xiàng),a2+b2ab(a+b)22ab(a+b)2a+b=a+b,當(dāng)且僅當(dāng)a=b時(shí)取等號(hào).故選D.答案D5.若lg x+lg y=2,則1x+1y的最小值為()A.120B.C.D.2解析由lg x+lg y=2可知x>0,y>0,且xy=100,于是1x+1y=x+yxy=1100(x+y)11002xy=15,當(dāng)且僅當(dāng)x=y=10時(shí),取等號(hào).故1x+1y的最小值為.答案B6.已知a>1,且m=loga(a2+1),n=loga(a+1),p=loga(2a),則m,n,p的大小關(guān)系是.(用“>”連接)解析a>1,a2+1>2a>a+1,loga(a2+1)>loga(2a)>loga(a+1),m>p>n.答案m>p>n7.已知t>0,則y=t2-3t+1t的最小值為.解析y=t2-3t+1t=t+-32t1t-3=-1,當(dāng)且僅當(dāng)t=1時(shí),取等號(hào).故函數(shù)的最小值為-1.答案-18.已知a>b>c,則(a-b)(b-c)與a-c2的大小關(guān)系是.解析a>b>c,a-b>0,b-c>0,a-c2=(a-b)+(b-c)2(a-b)(b-c).當(dāng)且僅當(dāng)b=a+c2時(shí)取等號(hào).答案(a-b)(b-c)a-c29.已知a,b均為正實(shí)數(shù),求證:1a2+1b2+ab22.證明由于a,b均為正實(shí)數(shù),所以1a2+1b221a21b2=2ab,當(dāng)且僅當(dāng)1a2=1b2,即a=b時(shí),等號(hào)成立.又因?yàn)?ab+ab22abab=22,當(dāng)且僅當(dāng)2ab=ab時(shí)等號(hào)成立,所以1a2+1b2+ab2ab+ab22,當(dāng)且僅當(dāng)1a2=1b2,2ab=ab,即a=b=42時(shí)取等號(hào).10.導(dǎo)學(xué)號(hào)04994085已知不等式ax2-3x+2<0的解集為A=x|1<x<b.(1)求a,b的值;(2)求函數(shù)f(x)=(2a+b)x-9(a-b)x(xA)的最小值.解(1)由題意知1+b=3a,1b=2a,a>0,解得a=1,b=2.(2)由(1)知a=1,b=2,A=x|1<x<2,所以f(x)=4x+ (1<x<2),而當(dāng)x>0時(shí),4x+24x9x=26=12.當(dāng)且僅當(dāng)4x=,即x=時(shí)取等號(hào).而x=32A,故f(x)的最小值為12.B組1.已知3a+2b=2(a>0,b>0),則ab的最小值是()A.4B.5C.6D.7解析3a+2b=2(a>0,b>0),223a2b,化為ab6,當(dāng)且僅當(dāng)a=3,b=2時(shí)取等號(hào).ab的最小值是6.故選C.答案C2.若a,b,cR,且ab+bc+ca=1,則下列不等式成立的是()A.a2+b2+c22B.a+b+c3C.1a+1b+1c23D.(a+b+c)23解析因?yàn)閍2+b22ab,b2+c22bc,a2+c22ac,于是a2+b2+c2ab+bc+ca=1,故A錯(cuò);而(a+b+c)2=a2+b2+c2+2(ab+bc+ca)3(ab+bc+ca)=3,故選項(xiàng)D正確;從而選項(xiàng)B錯(cuò)誤;令a=b=c=33,則ab+bc+ca=1,但1a+1b+1c=33>23,故選項(xiàng)C錯(cuò)誤.答案D3.已知x,y均為正數(shù),且xy,則下列四個(gè)數(shù)中最大的一個(gè)是()A.121x+1yB.1x+yC.1xyD.12(x2+y2)解析取x=1,y=2,可得121x+1y=34,1x+y=13,1xy=12,12(x2+y2)=110,因此最大的是121x+1y.答案A4.函數(shù)f(x)=x+4x的最小值等于.解析由基本不等式可知f(x)=x+4x2x4x=4,當(dāng)且僅當(dāng)x=4x,即x=4時(shí)取最小值.答案45.已知a>0,b>0,若lg a和lg b的等差中項(xiàng)是0,則1a+1b的最小值是.解析由已知得lg a+lg b=0,即ab=1,于是1a+1b=a+bab=a+b2ab=2,當(dāng)且僅當(dāng)a=b=1時(shí),取等號(hào).故1a+1b的最小值是2.答案26.已知函數(shù)f(x)=4x+ (x>0,a>0)在x=3處取得最小值,則a=.解析由基本不等式,得4x+24xax=4a,當(dāng)且僅當(dāng)4x=,即x=a2時(shí),等號(hào)成立,即a2=3,a=36.答案367.若x,yR,且滿足(x2+y2+2)(x2+y2-1)-180.(1)求x2+y2的取值范圍;(2)求證:xy2.(1)解由(x2+y2)2+(x2+y2)-200,得(x2+y2+5)(x2+y2-4)0,因?yàn)閤2+y2+5>0,所以有0x2+y24.故x2+y2的取值范圍是0,4.(2)證明由(1)知x2+y24,所以xyx2+y2242=2,當(dāng)且僅當(dāng)x=y時(shí),取等號(hào).故xy2.8.導(dǎo)學(xué)號(hào)04994086已知a,b為正實(shí)數(shù),且1a+1b=22.(1)求a2+b2的最小值;(2)若(a-b)24(ab)3,求ab的值.解(1)a,b為正實(shí)數(shù),且1a+1b=22,1a+1b=2221ab,即ab (當(dāng)且僅當(dāng)a=b時(shí)等號(hào)成立).a2+b22ab2=1(當(dāng)且僅當(dāng)a=b時(shí)等號(hào)成立),a2+b2的最小值為1.(2)1a+1b=22,a+b=22ab.(a-b)24(ab)3,(a+b)2-4ab4(ab)3,即(22ab)2-4ab4(ab)3,即(ab)2-2ab+10,(ab-1)20.a,b為正實(shí)數(shù),ab=1.