2024高考總復(fù)習(xí)優(yōu)化設(shè)計(jì)一輪用書(shū)數(shù)學(xué)配北師版(適用于老高考新教材)課時(shí)規(guī)范練3　等式性質(zhì)與不等式性質(zhì)

課時(shí)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(mǎn)足a<b,則下列關(guān)系式一定成立的是()A.a2<b2 B.ln(b-a)>0C.1a>1b D.22.(2021廣東高三二模)已知a,b∈R,且滿(mǎn)足ab<0,a+b>0,a>b,則()A.1a<1b BC.a2>b2 D.a<|b|3.(2021遼寧錦州高三期中)已知a>b,c>d,則下列關(guān)系式正確的是()A.ac+bd>ad+bc B.ac+bd<ad+bcC.ac>bd D.ac<bd4.(2021山西臨汾一中高三期中)已知1≤a+b≤5,-1≤a-b≤3,則3a-2b的取值范圍是()A.[-6,14] B.[-2,14]C.[-6,10] D.[-2,10]5.(2021浙江湖州高三月考)已知a,b,c∈(0,+∞),若ca+bA.c<a<b B.b<c<aC.a<b<c D.c<b<a6.(2021天津高三一模)已知x>0,y>0,lnyx>lgxy,則(A.1x>1y B.C.yx<xy D.7.(2021廣東實(shí)驗(yàn)中學(xué)高三模擬)已知正數(shù)x,y,z滿(mǎn)足xlny=yez=zx,則x,y,z的大小關(guān)系為()A.x>y>z B.y>x>zC.x>z>y D.z>y>x8.下列說(shuō)法錯(cuò)誤的是()A.若a<b<0,則a|a|<b|b|B.若a>0,b>0,c>0,則aC.若a>0,b>0,則a+ba+D.若a>0,b∈R,則a≥2b-b9.已知a>b>0,且a3-b3=3(a-b),則以下結(jié)論錯(cuò)誤的是()A.a>1 B.ab<1C.a+b<2 D.logab+logba>2綜合提升組10.(2021湖南師大附中高三期中)已知-1≤x+y≤1,1≤x-y≤3,則8x·14y的取值范圍是()A.[4,128] B.[8,256]C.[4,256] D.[32,1024]11.已知a,b∈R,則“|a-b|>|b|”是“ba<12A.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1≤x≤3≤y≤5,則()A.4≤x+y≤8B.x+y+1xC.-2≤x-y≤0D.x+1yy+4x的最小值為913.(2021湖北荊門(mén)高三期中)正實(shí)數(shù)a,b,c滿(mǎn)足1a+1b=1,1a+b創(chuàng)新應(yīng)用組14.(2021湖南岳陽(yáng)高三期中)已知2<x<4,-3<y<-1,則xx-2A.110,14 B.1C.15,1 D.23,215.已知a,b均為正數(shù),且a-b=1,則()A.2a-2b>1 B.a3-b3<1C.4a?1D.2log2a-log2b<2

課時(shí)規(guī)范練3等式性質(zhì)與不等式性質(zhì)1.D解析:對(duì)于A,a=-3,b=2滿(mǎn)足a<b,但是a2=9,b2=4,所以a2>b2,故A錯(cuò)誤;對(duì)于B,a=1,b=32滿(mǎn)足a<b,但是b-a=12,所以ln(b-a)<0,故B錯(cuò)誤;對(duì)于C,a=-3,b=2滿(mǎn)足a<b,但是1a=-13,1b=12,所以1a<1b,故C錯(cuò)誤;對(duì)于D,因?yàn)楹瘮?shù)y=2x在R上單調(diào)遞增,且a<b,2.C解析:因?yàn)閍b<0,a>b,所以a>0,b<0,1a>0,1b<0,故A不正確;ba<0,ab<0,則ba+ab<0,故B不正確;又a+b>0,即a>-b>0,所以a2>(-b)2,即a2>b2,故C正確;由a>-b>0得a>|b|3.A解析:∵a>b,c>d,∴ac+bd-(ad+bc)=(a-b)(c-d)>0,故A正確,B錯(cuò)誤;對(duì)于C,當(dāng)b=0,c<0時(shí),ac<0,bd=0,故C錯(cuò)誤;對(duì)于D,當(dāng)a>b>0,c>d>0時(shí),ac>bd,故D錯(cuò)誤.故選A.4.D解析:令3a-2b=m(a+b)+n(a-b)(m,n∈R),則m+n=3,m-n=-2,解得m=12,n=52.又因?yàn)?≤a+b≤5,-1≤a-b≤3,所以12≤12(5.A解析:由ca+b<ab+c<bc+a可得ca+b+1<ab+c+1<bc+6.A解析:∵lnyx>lgxy,∴l(xiāng)ny-lnx>lgx-lgy,∴l(xiāng)ny+lgy>lnx+lgx,∴y>x>0(函數(shù)y=lnx+lgx為增函數(shù)).對(duì)于A,y>x>0?1x>1y,故正確;對(duì)于B,取y=π,x=π2,siny=0<sinx=1,故錯(cuò)誤;對(duì)于C,取y=2,x=1,顯然不成立,故錯(cuò)誤;對(duì)于D,假設(shè)eyx>10xy成立,則lneyx>ln10xy,即yx>xyln10,可得y2>x2ln7.A解析:由xlny=zx,得z=lny,即y=ez,令f(z)=ez-z(z>0),則f'(z)=ez-1>0,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ez>z,即y>z.由yez=zx,得ez·ez=zx,即x=e2zz,所以x-y=e2zz-ez=e2z-zez8.B解析:對(duì)于A,由a<b<0,得a|a|=-a2,b|b|=-b2,且a2>b2,則-a2<-b2,即a|a|<b|b|,正確;對(duì)于B,a+cb+c?ab=ab+bc-ab-acb(b+c)=c(b-a)b(b+c),顯然當(dāng)b<a時(shí),ab>a+cb+c,錯(cuò)誤;對(duì)于C,由a>0,b>0,則a+ba+4ab=a2+a9.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a>b>0,∴a2+a2+a2>a2+ab+b2=3,即a2>1,a>1,故A正確;∵a2+b2≥2ab,當(dāng)且僅當(dāng)a=b時(shí),等號(hào)成立,∴a2+b2>2ab,則a2+ab+b2>3ab,即ab<1,故B正確;∵(a+b)2=a2+2ab+b2=3+ab<4,∴a+b<2,故C正確;∵a>1,ab<1,∴0<b<1,則logab<0,logba<0,則logab+logba<0,故D錯(cuò)誤.10.C解析:8x·14y=23x-2y.設(shè)3x-2y=m(x+y)-n(x-y)=(m-n)x+(m+n)y(m,n∈R),則有m-n=3,m+n=-2,解得m=12,n=-52.故3x-2y=12(x+y)+52(x-y).因?yàn)?1≤x+y≤1,1≤x-y≤3,所以3x-2y=12(x+y)+52(x-y11.C解析:由|a-b|>|b|得a2+b2-2ab>b2,∴a(a-2b)>0,∴a-2ba>0,∴1-2ba>0,∴ba<12.反之,也成立.故“|a-b|>|b|12.A解析:因?yàn)?≤x≤3≤y≤5,所以4≤x+y≤8,-4≤x-y≤0,故A正確,C錯(cuò)誤;因?yàn)閤+y+1x+16y=x+1x+y+16y≥2x·1x+2y·16y=10,當(dāng)且僅當(dāng)x=1,y=4時(shí),等號(hào)成立,所以x+y+1x+16y的最小值為10,故B錯(cuò)誤;因?yàn)閤+1yy+4x=xy+4xy+5≥24+5=9,當(dāng)且僅當(dāng)xy=2時(shí),等號(hào)成立,但1≤x≤3≤y≤5,xy取不到2,所以x+113.1,43解析:因?yàn)檎龑?shí)數(shù)a,b,c滿(mǎn)足1a+1b=1,1a+b+1c=1,所以c>1.又(a+b)1a+1b=2+ba+ab≥2+2ba·ab=4,當(dāng)且僅當(dāng)a=b=2時(shí),等號(hào)成立,所以a+b≥14.B解析:xx-2y=11-2yx,由已知得2<-2y<6,所以24<-2yx<62,即1215.A解析:對(duì)于A,因?yàn)閍-b=1,所以2a-2b=2b+1-2b=2b(2-1)=2b>1,故A正確;對(duì)于B,a3-b3=(a-b)(a2+ab+b2)=a2+ab+b2=(b+1)2+(b+1)