Đáp án:
$B=\dfrac{\sqrt{x}+2}{2\sqrt{x}+1}$
Giải thích các bước giải:
ĐKXĐ: $x\ge 0;\,x\ne \dfrac{1}{4}$
$B=\dfrac{1}{2\sqrt{x}-1}-\dfrac{2x+\sqrt{x}-3}{1-4x}$
$=\dfrac{2\sqrt{x}+1}{(2\sqrt{x}-1)(2\sqrt{x}+1)}+\dfrac{2x+\sqrt{x}-3}{4x-1}$
$=\dfrac{2\sqrt{x}+1+2x+\sqrt{x}-3}{4x-1}$
$=\dfrac{2x+3\sqrt{x}-2}{4x-1}$
$=\dfrac{(2\sqrt{x}-1)(\sqrt{x}+2)}{(2\sqrt{x}-1)(2\sqrt{x}+1)}$
$=\dfrac{\sqrt{x}+2}{2\sqrt{x}+1}$.
