課時規(guī)范練3　等式性質(zhì)與不等式性質(zhì)

1、 課時規(guī)范練3等式性質(zhì)與不等式性質(zhì)基礎(chǔ)鞏固組1.(2021河南鄭州高三月考)已知實(shí)數(shù)a,b滿足ab,則下列關(guān)系式一定成立的是()A.a20C.1a1bD.2a2b2.(2021廣東高三二模)已知a,bR,且滿足ab0,ab,則()A.1a0C.a2b2D.ab,cd,則下列關(guān)系式正確的是()A.ac+bdad+bcB.ac+bdbdD.acbd4.(2021山西臨汾一中高三期中)已知1a+b5,-1a-b3,則3a-2b的取值范圍是()A.-6,14B.-2,14C.-6,10D.-2,105.(2021浙江湖州高三月考)已知a,b,c(0,+),若ca+bab+cbc+a,則有()A.cab

2、B.bcaC.abcD.cb0,y0,lnyxlgxy,則()A.1x1yB.sin ysin xC.yx10 xy7.(2021廣東實(shí)驗(yàn)中學(xué)高三模擬)已知正數(shù)x,y,z滿足xln y=yez=zx,則x,y,z的大小關(guān)系為()A.xyzB.yxzC.xzyD.zyx8.下列說法錯誤的是()A.若ab0,則a|a|0,b0,c0,則ab0,b0,則a+ba+4ab4D.若a0,bR,則a2b-b2a9.已知ab0,且a3-b3=3(a-b),則以下結(jié)論錯誤的是()A.a1B.ab1C.a+b2綜合提升組10.(2021湖南師大附中高三期中)已知-1x+y1,1x-y3,則8x14y的取值范圍是

3、()A.4,128B.8,256C.4,256D.32,1 02411.已知a,bR,則“|a-b|b|”是“ba12”的()A.充分不必要條件B.必要不充分條件C.充要條件D.既不充分也不必要條件12.若1x3y5,則()A.4x+y8B.x+y+1x+16y取不到最小值C.-2x-y0D.x+1yy+4x的最小值為913.(2021湖北荊門高三期中)正實(shí)數(shù)a,b,c滿足1a+1b=1,1a+b+1c=1,則實(shí)數(shù)c的取值范圍是.創(chuàng)新應(yīng)用組14.(2021湖南岳陽高三期中)已知2x4,-3y1B.a3-b31C.4a-1b1D.2log2a-log2b2課時規(guī)范練3等式性質(zhì)與不等式性質(zhì)1.D解

4、析:對于A,a=-3,b=2滿足ab2,故A錯誤;對于B,a=1,b=32滿足ab,但是b-a=12,所以ln(b-a)0,故B錯誤;對于C,a=-3,b=2滿足ab,但是1a=-13,1b=12,所以1a1b,故C錯誤;對于D,因?yàn)楹瘮?shù)y=2x在R上單調(diào)遞增,且ab,所以2a2b,故D正確.故選D.2.C解析:因?yàn)閍bb,所以a0,b0,1b0,故A不正確;ba0,ab0,則ba+ab0,即a-b0,所以a2(-b)2,即a2b2,故C正確;由a-b0得a|b|,故D不正確.故選C.3.A解析:ab,cd,ac+bd-(ad+bc)=(a-b)(c-d)0,故A正確,B錯誤;對于C,當(dāng)b=0

5、,c0時,acb0,cd0時,acbd,故D錯誤.故選A.4.D解析:令3a-2b=m(a+b)+n(a-b)(m,nR),則m+n=3,m-n=-2,解得m=12,n=52.又因?yàn)?a+b5,-1a-b3,所以1212(a+b)52,-5252(a-b)152,故-23a-2b10.5.A解析:由ca+bab+cbc+a可得ca+b+1ab+c+1bc+a+1,即a+b+ca+ba+b+cb+cb+cc+a.由a+bb+c可得ac,由b+cc+a可得ba,于是有calgxy,ln y-ln xlg x-lg y,ln y+lg yln x+lg x,yx0(函數(shù)y=ln x+lg x為增函數(shù)

6、).對于A,yx01x1y,故正確;對于B,取y=,x=2,sin y=010 xy成立,則ln eyxln 10 xy,即yxxyln 10,可得y2x2ln 10,而當(dāng)yx0時,不能一定有y2x2ln 10,故不成立.故選A.7.A解析:由xln y=zx,得z=ln y,即y=ez,令f(z)=ez-z(z0),則f(z)=ez-10,所以函數(shù)f(z)在(0,+)上單調(diào)遞增,所以f(z)f(0)=e0-0=1,所以ezz,即yz.由yez=zx,得ezez=zx,即x=e2zz,所以x-y=e2zz-ez=e2z-zezz=ez(ez-z)z0,所以xy.綜上,xyz,故選A.8.B解析

7、:對于A,由abb2,則-a2-b2,即a|a|b|b|,正確;對于B,a+cb+c-ab=ab+bc-ab-acb(b+c)=c(b-a)b(b+c),顯然當(dāng)ba+cb+c,錯誤;對于C,由a0,b0,則a+ba+4ab=a2+a2+ba+4ab44a2a2ba4ab=4,當(dāng)且僅當(dāng)a2=ba=4ab,即a=b=2時,等號成立,正確;對于D,a0,bR,而(a-b)20,即a22ab-b2,故a2b-b2a,正確.故選B.9.D解析:由立方差公式可得a3-b3=(a-b)(a2+ab+b2)=3(a-b),則a2+ab+b2=3,又ab0,a2+a2+a2a2+ab+b2=3,即a21,a1,

8、故A正確;a2+b22ab,當(dāng)且僅當(dāng)a=b時,等號成立,a2+b22ab,則a2+ab+b23ab,即ab1,故B正確;(a+b)2=a2+2ab+b2=3+ab4,a+b1,ab1,0b1,則logab0,logba0,則logab+logba|b|得a2+b2-2abb2,a(a-2b)0,a-2ba0,1-2ba0,ba|b|”是“ba1.又(a+b)1a+1b=2+ba+ab2+2baab=4,當(dāng)且僅當(dāng)a=b=2時,等號成立,所以a+b4,則01a+b14,即01-1c14,解得1c43.14.B解析:xx-2y=11-2yx,由已知得2-2y6,所以24-2yx62,即12-2yx3,所以321-2yx4,所以1411-2yx1,故A正確;對于B,a3-b3=(a-b)(a2+ab+b2)=a2+ab+b2=(b+1)2+(b+1)b+b2=3b2+3b+11,故B錯誤;對于C,4a-1b=4a-1b(a-b)=4+1-ab-4ba=5-ab+4ba5-2ab4ba=1,當(dāng)