Giải phương trình c) 2/x+3 - 3/x-2= 1-5x/(x+3)(x-2)

c) ĐKXĐ: $\left\{\begin{array}{c}x+3\ne 0\\ x-2\ne 0\end{array}\right.\iff \left\{\begin{array}{c}x\ne -3\\ x\ne 2\end{array}\right.$

Ta có: $\frac{2}{x+3}-\frac{3}{x-2}=\frac{1-5x}{\left(x+3\right)\left(x-2\right)}$

$\iff \frac{2\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{3\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{1-5x}{\left(x+3\right)\left(x-2\right)}$

$\iff \frac{2x-4}{\left(x+3\right)\left(x-2\right)}-\frac{3x+9}{\left(x+3\right)\left(x-2\right)}=\frac{1-5x}{\left(x+3\right)\left(x-2\right)}$

$\iff \frac{2x-4-3x-9}{\left(x+3\right)\left(x-2\right)}=\frac{1-5x}{\left(x+3\right)\left(x-2\right)}$

$\iff \frac{-x-13}{\left(x+3\right)\left(x-2\right)}=\frac{1-5x}{\left(x+3\right)\left(x-2\right)}$

Þ -x - 13 = 1 - 5x

Û -x + 5x = 1 + 13

Û 4x = 14

$\iff x=\frac{14}{4}=\frac{7}{2}$ (thỏa mãn ĐKXĐ).

Vậy tập nghiệm của phương trình là $S=\left\{\frac{7}{2}\right\}.$