不等式的解法举例练习一、选择题1.下列不等式中与0)2lg(x同解的是()(A)0)2)(3(xx(B)023xx(C)032xx(D)0)2)(3(xx2.不等式1)2(logxx的解集是()(A)(),2((B)),1((C)(0,1)(D)(0,1)),1(3.不等式022bxax的解集是)31,21(,则a+b的值是().(A)10(B)-10(C)14(D)-144.不等式1652xxx的解集是().(A))1,((B)),2((C)35,1(D))35,(5.已知集合M}22|{322xxx,N}0)1(log|{21xx则MN=().(A))23,0((B))2,32((C))23,1((D)(0,1)二、填空题6.不等式22214xaxax对一切xR恒成立,则实数a的取值范围是_______.7.函数)12(log1.0xy的定义域是_____________.8.不等式043)4(2xxx的解集是____________.9.若关于x的不等式02baxx的解集是}32|{xx,则关于x的不等式012axbx的解集是_____________.10.若关于x的不等式xxkkxk122)232()232(的解集是),21(,则实数k的取值范围是________.三、解答题11.解关于x的不等式1log22log3xxaa(0a,且1a).12.解关于x的不等式:3)93(log)13(log233xx.13.已知x满足:03log7)(log221221xx,求)4(log)2(log)(22xxxf的最大值和最小值.14.对一切xR,不等式12sin23cos2mxxm恒成立,求实数m的取值范围.15.二次函数)0()(2acbxaxxf对一切xR都有)2()2(xfxf,解不等式)852(log)21(log221221xxfxxf.参考答案1.A2.B3.D4.D5.C6.(2,)7.(0,1)8.{-1}[4,)9.31(,)2110.221221k11.212log3012log33)2log3(22xxxaaa或2log3xa43log321xa或1logxa.若a>1,则不等式的解集为32[a,[]43aa,);若0<a<1,则不等式的解集是43[a,(]32a0,a].12.1)13(log303)13(log2)]13([log3323xxx313271x4log328log43272833xx.13.先求得3log212x.把f(x)整理,得:41)23(log)(22xxf,23log)()(2maxxxfxf,4123log)()(2minxxfxf.14.32621m.15.∵241)21(log)21(log221221xxx,121)41(2log)852(log221221xxx,又f(x)在(,2]上递增,由原不等式,得:)852(log)21(log221221xxxx8522108520212222xxxxxxxx4414141x.