`#DyHungg`
`2/(x-1)=1+(2x)/(x+2)`
`ĐKXĐ:x` $\neq$ `1;-2`
`⇔(2(x+2))/(x-1)-((x-1)(x+2))/((x-1)(x+2))-(2x(x-1))/((x-1)(x+2))=0`
`⇒2x+4-(x²+2x-x-2)-2x²+2x=0`
`⇔-3x²+3x+6=0`
`⇔-3x²+6x-3x+6=0`
`⇔-3x(x-2)-3(x-2)=0`
`⇔(-3x-3)(x-2)=0`
`1) -3x-3=0⇔x=-1 (TM)`
`2) x-2=0⇔x=2 (TM)`
Vậy.........