Toán $\frac{x+3}{x-2}$ – $\frac{x-10}{x^2-4}$ = $\frac{x+5}{x+2}$ 21/11/2021 By Cora $\frac{x+3}{x-2}$ – $\frac{x-10}{x^2-4}$ = $\frac{x+5}{x+2}$
Đáp án: Giải thích các bước giải: $\dfrac{x+3}{x-2}-\dfrac{x-10}{x^2-4}=\dfrac{x+5}{x+2}$ $ĐKXĐ:x\neq-2;2$ $=>(x+3)(x+2)-(x-10)-(x+5)(x-2)=0$ $=>x^2+5x+6-x+10-x^2-3x+10=0$ $=>x+26=0$ $=>x=-26$ Trả lời
$\frac{x+3}{x-2}-\frac{x-10}{x^2-4}=\frac{x+5}{x+2}$ $Đkxđ:x\neq±2$ $⇒x^2+2x+3x+6-x+10=x^2+5x-2x-10$ $⇔x=-26$ $(tm$ $đkxđ)$ Vậy $S=${$-26$}. Trả lời
Đáp án:
Giải thích các bước giải:
$\dfrac{x+3}{x-2}-\dfrac{x-10}{x^2-4}=\dfrac{x+5}{x+2}$ $ĐKXĐ:x\neq-2;2$
$=>(x+3)(x+2)-(x-10)-(x+5)(x-2)=0$
$=>x^2+5x+6-x+10-x^2-3x+10=0$
$=>x+26=0$
$=>x=-26$
$\frac{x+3}{x-2}-\frac{x-10}{x^2-4}=\frac{x+5}{x+2}$ $Đkxđ:x\neq±2$
$⇒x^2+2x+3x+6-x+10=x^2+5x-2x-10$
$⇔x=-26$ $(tm$ $đkxđ)$
Vậy $S=${$-26$}.