$\frac{1-x}{1+x}$ +3 =$\frac{2x+3}{x+1}$
⇔$\frac{1-x}{1+x}$ +$\frac{3+3x}{1+x}$= $\frac{2x+3}{x+1}$
⇒1+3+3x=2x+3
⇔3x-2x=3-1-3
⇔x=-1(KTMDK)
vậy x=∅
b,$\frac{13}{(x-3)(2x+7)}$+ $\frac{2}{2x+7}$ =$\frac{x}{x^{2}-9}$
⇔$\frac{13x+39}{(x+3)(x-3)(2x+7)}$+ $\frac{2x^{2}-18}{(x-3)(x+3)(2x+7)}$= $\frac{2x^{2}+7x}{(x-3)(x+3)(2x+7)}$
⇒13x + 39 +2$x^{2}$-18=2$x^{2}$+7x
⇔13x+2$x^{2}$-2$x^{2}$-7x=21
⇔6x=21
⇔x=$\frac{21}{6}$
Vậy x=$\frac{21}{6}$