Đáp án:
$ 82 ) \dfrac{x-4}{x+3} + \dfrac{5+2x}{3-x} = 1 - \dfrac{x^2}{x^2-9}$
$\text{ĐKXĐ : x $\neq$±3}$
$⇔ \dfrac{x-4}{x+3} +\dfrac{-5-2x}{x-3} = 1 -\dfrac{x^2}{x^2-9}$
$⇔\dfrac{(x-4)(x-3)}{(x+3)(x-3)} + \dfrac{(-5-2x)(x+3)}{(x-3)(x+3)} = \dfrac{(x-3)(x+3)}{(x-3)(x+3)} - \dfrac{x^2}{(x-3)(x+3)}$
$⇒ (x-4)(x-3) + (-5-2x)(x+3) = (x-3)(x+3) - x^2$
$⇔ x^2-3x-4x+12 -5x-15-2x^2-6x = x^2-9 -x^2$
$⇔x^2-2x^2-x^2+x^2 -3x-4x -5x -6x+12-15+9=0$
$⇔ -x^2-18x+6 =0$
$⇔ -(x^2+2.x.9 +81 - 87)=0$
$⇔-(x+9)^2 =-√87$
$⇔x = ±√87 - 9$(TM)
Vậy ....
