无理不等式的解法基本概念1 、无理不等式:2 、无理不等式的类型: 根号下含有未知数的不等式。0)()()4()()()3()()()2()()()1(xgxfxgxfxgxfxgxf根式不等式的解法 -------例 1 解不等式0343xx解:原不等式可化为343xx根据根式的意义及不等式的性质,得34303043xxxx解这个不等式组,得3|xx所以,原不等式的解集为3|xx34| xx 21| xx3|xx21343⊙⊙●)()(xgxf)()(0)(0)(xgxfxgxf)()(0)(xgxfxf根式不等式的解法 ------- 类型( 1 )根式不等式的解法 -------例 2 解不等式03227xx解:原不等式可化为3227xx根据根式的意义及不等式的性质,得)2...(032027)1...()3(270320272xxxxxx或解这个不等式组( 1 ),得923|92|23|27|xxxxxxxx所以,原不等式的解集为923|2327|xxxx解这个不等式组( 2 ),得2327|23|27|xxxxxx927|xx272239●●2723●根式不等式的解法 ------- 类型( 2 ))()(xgxf0)(0)()]([)(0)(0)(2xgxfxgxfxgxf或0)(0)([g(x)]f(x)0g(x)2xgxf或根式不等式的解法 -------例 3 解不等式03227xx解:原不等式可化为3227xx根据根式的意义及不等式的性质,得2)3(27032027xxxx解这个不等式组,得92|23|27|xxxxxxx或9|xx-27-23/29根式不等式的解法 ------- 类型( 3 ))()(xgxf2)]([)(0)(0)(xgxfxgxf根式不等式的解法 -------例 4 解不等式032)2(2xxx解:原不等式可化为0320320222xxxxx或解这个不等式组,得2|xx31|xxx或-2-3113|xxx或1,3●●根式不等式的解法 ------- 类型( 4 )0)()()1(xgxf0)(0)(xgxf0)()()2(xgxf0)(0)(0)(xfxgxf或根式不等式的解法练习 -------032)4(242)3(2)2(01341222xxxxxxxxxx)(解下列不等式:4|xx6|xx31|xxx或R答案:小结:)()(xgxf)()(0)(0)(xgxfxgxf)()(xgxf0)(0)()]([)(0)(0)(2xgxfxgxfxgxf或)()(xgxf2)]([)(0)(0)(xgxfxgxf1.2.3.4.作业:02)32(722924xxPP解不等式补充题:习题十六:练习: