Giải phương trình d) 5x/x-3 + 2/x+3 = x^2+5x-6/ x^2 -9

d) ĐKXĐ: $\left\{\begin{array}{c}x-3\ne 0\\ x+3\ne 0\end{array}\right.\iff \left\{\begin{array}{c}x\ne 3\\ x\ne -3\end{array}\right.$

$\frac{5x}{x-3}+\frac{2}{x+3}=\frac{x^2+5x-6}{x^2-9}$

$\iff \frac{5x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{x^2+5x-6}{\left(x-3\right)\left(x+3\right)}$

$\iff \frac{5x^2+15x}{\left(x-3\right)\left(x+3\right)}+\frac{2x-6}{\left(x-3\right)\left(x+3\right)}=\frac{x^2+5x-6}{\left(x-3\right)\left(x+3\right)}$

$\iff \frac{5x^2+17x-6}{\left(x-3\right)\left(x+3\right)}=\frac{x^2+5x-6}{\left(x-3\right)\left(x+3\right)}$

Þ 5x² + 17x - 6 = x² + 5x - 6

Û 4x² + 12x = 0

Û 4x(x + 3) = 0

$\Rightarrow \left[\begin{array}{c}x=0\\ x+3=0\end{array}\right.$

$\iff \left[\begin{array}{c}x=0\\ x=-3\end{array}\right.$

Đối chiếu ĐKXĐ suy ra x = 0 thỏa mãn

Vậy nghiệm của phương trình là x = 0.