$\frac{2x-3}{x+2}$ + $\frac{x-5}{3-x}$ – $\frac{10}{x^2-x-6}$ = 2 giải phương trình

`frac{2x-3}{x+2}+frac{x-5}{3-x}-frac{10}{x^2-x-6}=2`   ĐKXĐ: `x\ne-2;x\ne3`

`<=>frac{2x-3}{x+2}+frac{x-5}{-(x-3)}-frac{10}{x^2+2x-3x-6}=2`

`<=>frac{2x-3}{x+2}-frac{x-5}{x-3}-frac{10}{(x+2)(x-3)}=2`

`<=>frac{(2x-3)(x-3)}{(x+2)(x-3)}-frac{(x-5)(x+2)}{(x-3)(x+2)}-frac{10}{(x+2)(x-3)}=frac{2(x-3)(x+2)}{(x-3)(x+2)}`

`=>(2x-3)(x-3)-(x-5)(x+2)-10=2(x-3)(x+2)`

`<=>2x^2-6x-3x+9-(x^2+2x-5x-10)-10=(2x-6)(x+2)`

`<=>2x^2-9x+9-x^2+3x+10-10=2x^2+4x-6x-12`

`<=>x^2-6x+9=2x^2-2x-12`

`<=>x^2-6x+9-2x^2+2x+12=0`

`<=>-x^2-4x+21=0`

`<=>-(x^2+4x-21)=0`

`<=>x^2+4x-21=0`

`<=>x^2+7x-3x-21=0`

`<=>(x^2+7x)-(3x+21)=0`

`<=>x(x+7)-3(x+7)=0`

`<=>(x+7)(x-3)=0`

`<=>` \(\left[ \begin{array}{l}x+7=0\\x-3=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-7(TMĐK)\\x=3(KTMĐK)\end{array} \right.\) 

Vậy phương trình trên có tập nghiệm `S={-7}`