Giải thích các bước giải:
đkxđ $x\ne 5,-6, -5,6$
$\dfrac{x+6}{x-5}+\dfrac{x-5}{x+6}=\dfrac{2x^2+23x+61}{x^2-x-30}$
$\to \dfrac{x+6}{x-5}+\dfrac{x-5}{x+6}=\dfrac{2x^2+23x+61}{(x+5)(x-6)}$
$\to (x+6)^2(x+5)(x-6)+(x-5)^2(x+5)(x-6)=(2x^2+23x+61)(x+6)(x-5)$
$\to 25x^3+25x^2-508x=0$
$\to x(25x^2+25x-508)=0$
$\to x\in\{0,\dfrac{-5+11\sqrt{17}}{10},\dfrac{-5-11\sqrt{17}}{10}\}$
