Đáp án:
Giải thích các bước giải:
ĐKXĐ : x≠{8,9,10,11}
$\frac{8}{x-8}$+ $\frac{11}{x-11}$= $\frac{9}{x-9}$+ $\frac{10}{x-10}$
⇔$\frac{8}{x-8}$+1+ $\frac{11}{x-11}$+1- $\frac{9}{x-9}$+1- $\frac{10}{x-10}$+1 =0
⇔$\frac{8+x-8}{x-8}$+ $\frac{11+x-11}{x-11}$-$\frac{9+x-9}{x-9}$- $\frac{10+x-10}{x-10}$=0
⇔$\frac{x}{x-8}$+ $\frac{x}{x-11}$- $\frac{x}{x-9}$- $\frac{x}{x-10}$=0
⇔x($\frac{1}{x-8}$+ $\frac{1}{x-11}$- $\frac{1}{x-9}$- $\frac{1}{x-10}$)=0
⇔x=0
Vậy x=0
