Đáp án: x = 0,5
Giải thích các bước giải:
ĐKXĐ: x$\neq${-1;2}
Ta có:
$\frac{x+2}{x+1}$ - $\frac{3}{2-x}$ = $\frac{3}{x^2-x-2}$ + 1
⇔ $\frac{(x+2).(x-2)}{(x+1).(x-2)}$ + $\frac{3.(x+1)}{(x-2).(x+1)}$ = $\frac{3}{(x-2).(x+1)}$ + $\frac{(x-2).(x+1)}{(x-2).(x+1)}$
⇔ $\frac{x^2-4}{(x+1).(x-2)}$ + $\frac{3x+3}{(x-2).(x+1)}$ = $\frac{3}{(x-2).(x+1)}$ + $\frac{x^2-x-2}{(x-2).(x+1)}$
⇔ $\frac{x^2-4+3x+3}{(x+1).(x-2)}$ = $\frac{3+x^2-x-2}{(x-2).(x+1)}$
⇔ $x^2 + 3x-1$ = $x^2-x+1$
⇔ 4x = 2
⇔ x = 0,5 (thỏa mãn)
