a)$\frac{7}{x-3}$ -$\frac{5x-1}{x+3}$ = $\frac{87-5x^2}{x^2-9}$
⇔$\frac{7}{x-3}$ -$\frac{5x-1}{x+3}$ = $\frac{87-5x^2}{x^2-3^2}$
⇔$\frac{7}{x-3}$ -$\frac{5x-1}{x+3}$ = $\frac{87-5x^2}{(x+3)(x-3)}$
MTC:(x+3)(x-3)
ĐKXĐ:x`ne`-3 hoặc x `ne`3
⇔$\frac{7(x+3)}{(x+3)(x-3)}$ -$\frac{(5x-1)(x-3)}{(x+3)(x-3)}$ = $\frac{87-5x^2}{(x+3)(x-3)}$
⇒$7x+21-(5x^2-15x-x+3)=87-5x^2$
⇔$7x+21-5x^2+15x+x-3=87-5x^2$
⇔$7x+21-5x^2+15x+x-3-87+5x^2=0$
⇔$23x-69=0$
⇔$23(x-3)=0$
⇒$x-3=0$
⇔$x=3(loại)$
Vậy S=∅
b)$\frac{3x}{x-2}$ - $\frac{20}{x^2-4}$ = $\frac{5}{x+2}$
⇔$\frac{3x}{x-2}$ - $\frac{20}{x^2-2^2}$ = $\frac{5}{x+2}$
⇔$\frac{3x}{x-2}$ - $\frac{20}{(x-2)(x+2)}$ = $\frac{5}{x+2}$
MTC:(x-2)(x+2)
ĐKXĐ:x `ne` 2 hoặc x `ne`-2
⇔$\frac{3x(x+2)}{(x-2)(x+2)}$ - $\frac{20}{(x-2)(x+2)}$ = $\frac{5(x-2)}{(x-2)(x+2)}$
⇒$3x^2+6x-20=5x-10$
⇔$3x^2+6x-20-5x+10=0$
⇔$3x^2+x-10=0$
⇔$3x^2+6x-5x-10=0$
⇔$3x(x+2)-5(x+2) =0$
⇔$(x+2)(3x-5) =0$
⇒ $ x+2=0$ hoặc $3x-5=0$
$x=-2 (loại) $ hoặc $ 3x=5$
$x=-2 (loại)$ hoặc $x=\frac{5}{3} (nhận)$
Vậy S={$\frac{5}{3}$}
