Giải các phương trình b) 4/ x+2 + 3/x-2 = x+1/x^2 -4

b)$\frac{4}{x+2}+\frac{3}{x-2}=\frac{x+1}{x^2-4}$

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

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

Khi đó phương trình đã cho trở thành:

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

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

$\iff \frac{4x-8}{\left(x-2\right)\left(x+2\right)}+\frac{3x+6}{\left(x-2\right)\left(x+2\right)}=\frac{x+1}{\left(x-2\right)\left(x+2\right)}$

$\iff \frac{4x-8+3x+6}{\left(x-2\right)\left(x+2\right)}=\frac{x+1}{\left(x-2\right)\left(x+2\right)}$

$\iff \frac{7x-2}{\left(x-2\right)\left(x+2\right)}=\frac{x+1}{\left(x-2\right)\left(x+2\right)}$

Þ 7x -2 = x + 1

Û 7x - x = 2 + 1

Û 6x = 3  (TMĐK)

Vậy tập nghiệm của phương trình là $S=\left\{\frac{1}{2}\right\}.$