Giải thích các bước giải:
a.ĐKXĐ: $x\ne\pm\dfrac12$
Ta có:
$\dfrac{2}{2x+1}+\dfrac{x}{4x^2-1}=\dfrac{7}{2x-1}$
$\to \dfrac{2}{2x+1}+\dfrac{x}{(2x-1)(2x+1)}=\dfrac{7}{2x-1}$
$\to 2\left(2x-1\right)+x=7\left(2x+1\right)$
$\to 5x-2=14x+7$
$\to 9x=-9$
$\to x=-1$
b.ĐKXĐ: $x\ne\pm5$
Ta có:
$\dfrac{x^2+5}{25-x^2}=\dfrac{3}{x+5}+\dfrac{x}{x-5}$
$\to -\dfrac{x^2+5}{x^2-25}=\dfrac{3}{x+5}+\dfrac{x}{x-5}$
$\to -\dfrac{x^2+5}{x^2-5^2}=\dfrac{3}{x+5}+\dfrac{x}{x-5}$
$\to -\dfrac{x^2+5}{(x-5)(x+5)}=\dfrac{3}{x+5}+\dfrac{x}{x-5}$
$\to -(x^2+5)=3(x-5)+x(x+5)$
$\to -x^2-5=3x-15+x^2+5x$
$\to -x^2-5=x^2+8x-15$
$\to 2x^2+8x-10=0$
$\to x^2+4x-5=0$
$\to x^2+4x+4-9=0$
$\to (x+2)^2=9$
$\to x+2=3\to x=1$ hoặc $x+2=-3\to x=-5$
