`#Dark`
`a)(2x-5)/(x+5)=3`
`ĐKXĐ:x`$\neq$`-5`
`⇔ (2x-5)(x+5)=(3(x+5))/(x+5)`
`⇒2x-5=3x+15`
`⇔2x-3x=5+15`
`⇔-x=20`
`⇔x=-20(TM)`
Vậy `S={-20}`
`b)x/(x-1)=(x+4)/(x+1)`
`ĐKXĐ:x`$\neq$`±1`
`⇔(x(x+1))/((x-1)(x+1))=((x-1)(x+4))/((x-1)(x+1))`
`⇒x²+x=(x-1)(x+4)`
`⇔x²+x =x²+4x-x-4`
`⇔x²-x²+x+x-4x=-4`
`⇔-2x=-4`
`⇔x=2(TM)`
Vậy `S={2}`
`c)3/(x-2)=(2x-1)/(x-2)-x`
`ĐKXĐ:x`$\neq$`±2`
`⇔3/(x-2)=(2x-1)/(x-2)-(x(x-2))/(x-2)`
`⇒3=2x-1-x²+2x`
`⇔2x-1-x²+2x=3`
`⇔-x²+4x-4=0`
`⇔-x²+2x+2x-4=0`
`⇔-x(x-2)+2(x-2)=0`
`⇔(x-2)(-x+2)=0`
`1)x-2=0⇔x=2(KTM)`
`2)-x+2=0⇔x=2(KTM)`
Vậy pt vô nghiệm