怎么解这样的方程？1. (绝对值) |x-4/x| < 3 2. (x-2)·(10-x)/(x-3)≤03. x+1+5x２/(x+2)≥1/(1-x)要过程,谢谢
热心网友
解:由|x-4/x| < 3 得:(x-4/x)^< 9即:(x^-4)^<9x^∴(x^)-17x^+16< 0∴1< x^< 16∴-4< x<-1或1< x< 4(x-2)(x-10)/(x-3)≥0方程(x-2)(x-10)(x-3)=0的三根由小到大排列是2,3,10.由根轴法:2≤x＜3或x≥10[注意:x≠3]由x+1+5x/(x+2)-1/(1-x)≥0[x(x^-7x-5)]/(x+2)(x-1)≥0方程x(x^-7x-5)(x+2)(x-1)=0的根由小到大排列是-2,(7-√69)/2,0,1,(7+√69)/2.由根轴法:0＜x≤1或0≤x＜1或x≥(7+√69)/2[注意:x≠-2且x≠1]
热心网友
解:由|x-4/x| < 3 得:(x-4/x)^< 9即:(x^-4)^<9x^∴(x^)-17x^+16< 0∴1< x^< 16∴-4< x<-1或1< x< 4(x-2)(x-10)/(x-3)≥0方程(x-2)(x-10)(x-3)=0的三根由小到大排列是2,3,10.由根轴法:2≤x＜3或x≥10[注意:x≠3]由x+1+5x/(x+2)-1/(1-x)≥0[x(x^-7x-5)]/(x+2)(x-1)≥0方程x(x^-7x-5)(x+2)(x-1)=0的根由小到大排列是-2,(7-√69)/2,0,1,(7+√69)/2.由根轴法:0＜x≤1或0≤x＜1或x≥(7+√69)/2[注意:x≠-2且x≠1]