Giải thích các bước giải:
a.Ta có $(2x-5)(x+7)=0$
$\to 2x-5=0\to 2x=5\to x=\dfrac52$
Hoặc $x+7=0\to x=-7$
$\to x\in\{\dfrac52, -7\}$
b.ĐKXĐ: $x\ne \pm5$
Ta có:
$\dfrac{x+5}{x-5}-\dfrac{x-5}{x+5}=\dfrac{20}{x^2-25}$
$\to \dfrac{x+5}{x-5}-\dfrac{x-5}{x+5}=\dfrac{20}{(x+5)(x-5)}$
$\to (x+5)^2-(x-5)^2=20$
$\to (x^2+10x+25)- (x^2-10x+25)=20$
$\to 20x=20$
$\to x=1$
c.Ta có:
$(2x-3)^2-(x+1)^2=0$
$\to ((2x-3)- (x+1))((2x-3)+(x+1))=0$
$\to (x-4)(3x-2)=0$
$\to x\in\{4, \dfrac23\}$
d.ĐKXĐ: $x\ne -1, 3$
Ta có:
$\dfrac{x}{2(x-3)}+\dfrac{x}{2(x+1)}=\dfrac{2x}{(x+1)(x-3)}$
$\to x\left(x+1\right)+x\left(x-3\right)=4x$
$\to x(x+1+x-3)=4x$
$\to x(2x-2)=4x$
$\to x(x-1)=2x$
$\to x(x-1)-2x=0$
$\to x(x-1-2)=0$
$\to x(x-3)=0$
$\to x=0$ vì $x\ne 3$
