Giải thích các bước giải:
Câu 5:
a.$D=\dfrac{2}{x-3}-\dfrac{2x}{x^2-9}+\dfrac{x-1}{2x+6}$
$\rightarrow D=\dfrac{4(x+3)}{2(x+3)(x-3)}-\dfrac{4x}{2(x-3)(x+3)}+\dfrac{(x-1)(x-3)}{2(x+3)(x-3)}$
$\rightarrow D=\dfrac{4(x+3)-4x+(x-1)(x-3)}{2(x+3)(x-3)}$
$\rightarrow D=\dfrac{4x+12-4x+x^2-4x+3}{2(x+3)(x-3)}$
$\rightarrow D=\dfrac{x^2-4x+15}{2(x+3)(x-3)}$
b.Để $D=\dfrac{2}{3}$
$\rightarrow \dfrac{x^2-4x+15}{2(x+3)(x-3)}=\dfrac{2}{3}$
$\rightarrow 3x^2-12x+45=4(x^2-9)$
$\rightarrow x^2+12x-81=0$
$\rightarrow x=-6\pm 3\sqrt{13}$
