`a) 1/(x-1)+2/(x+1)=x/(x^2-1)` ĐKXĐ `x\ne ±1`
`⇔(x+1)/(x^2-1)+(2(x-1))/(x^2-1)=x/(x^2-1)`
`⇒x+1 + 2x-2=x`
`⇔x +2x-x=-1+2`
`⇔2x=1`
`⇔x=1/2` `(tm)`
`b)(2x-2x^2)/(x+3)=(4x)/(x+3)+2/7` ĐKXĐ `x\ne 3`
`⇔(7(2x-2x^2))/(7(x+3))=(28x)/(7(x+3))+(2(x+3))/(7(x+3)`
`⇒14x-14x^2=28x+2x+6`
`⇔14x-14x^2-28x-2x=6`
`⇔-14x^2-16x-6=0` (vô nghiệm )
Vậy pt vô nghiệm
`c)(2x-5)/(x+5)=3` ĐKXĐ `x\ne -5`
`⇒3(x+5)=2x-5`
`⇔3x+15=2x-5`
`⇔3x-2x=-5-15`
`⇔x=-20` `(tm)`
