`(x+1)/(x-2+x)-1/(x+2)=(2(x*2+2))/(x*2-4)`
`<=>(x+1)/(2(x-1))-1/(x+2)=(2(x+1))/(x-2)`
`<=>(x+1)(x+2)(x-2)-2(x-1)(x-2)=4(x+1)(x-1)(x+2)`
`<=>-3x^3-9x^2+6x=0`
`<=>x(x^2+3x-2)=0`
`<=>x=0` hoặc $\left \{ {{\frac{-3+\sqrt[]{17}}{2}} \atop {\frac{-3-\sqrt[]{17}}{2}}} \right.$
Vậy ...........