Đáp án:$\frac{3}{4x-20}$ +$\frac{15}{50-2x^{2}}$ +$\frac{7}{6x+30}$=0
Giải thích các bước giải:
$\frac{3}{4(x-5)}$ - $\frac{15}{(x-5)(x+5)}$ +$\frac{7}{6(x+5)}$ =0
$\frac{3.3(x+5)}{4.3(x-5)(x+5)}$ - $\frac{15.12}{12(x-5)(x+5)}$ +$\frac{7.2(x+5)}{6.2(x-5)(x+5)}$=0
$\frac{9(x+5)}{12(x-5)(x+5)}$ - $\frac{180}{12(x-5)(x+5)}$ +$\frac{14(x+5)}{12(x+5)(x+5)}$ =0
⇒9x+45-180+14x+70
⇔23x-65=0
⇔23x=65
⇔x=$\frac{65}{23}$
