Đáp án: a.$x=2$
b.$ x=5$
Giải thích các bước giải:
a.ĐKXĐ: $x\ne \pm3, \dfrac72$
Ta có:
$\dfrac{6}{x^2-9}-\dfrac{1}{2x-7}=\dfrac{13}{(x+3)(2x-7)}$
$\to \dfrac{6}{(x-3)(x+3)}-\dfrac{1}{2x-7}=\dfrac{13}{(x+3)(2x-7)}$
$\to 6\left(2x-7\right)-\left(x-3\right)\left(x+3\right)=13\left(x-3\right)$
$\to -x^2+12x-33=13x-39$
$\to x^2+x-6=0$
$\to (x+3)(x-2)=0$
Mà $x\ne -3\to x+3\ne 0\to x-2=0\to x=2$
b.ĐKXĐ: $x\ne -2,-3,-1$
Ta có:
$\dfrac{2(x-5)}{x^2+4x+3}=\dfrac{x-5}{x^2+5x+6}$
$\to \dfrac{2(x-5)}{(x+1)(x+3)}=\dfrac{x-5}{(x+2)(x+3)}$
$\to \dfrac{2(x-5)}{(x+1)(x+3)}-\dfrac{x-5}{(x+2)(x+3)}=0$
$\to \dfrac{x-5}{x+3}(\dfrac{2}{x+1}-\dfrac{1}{x+2})=0$
$\to x-5=0\to x=5$
Hoặc $\dfrac{2}{x+1}-\dfrac{1}{x+2}=0$
$\to \dfrac2{x+1}=\dfrac1{x+2}$
$\to 2(x+2)=x+1$
$\to 2x+4=x+1$
$\to x=-3$ loại vì $x\ne -3$
