Đáp án:
$B=-\dfrac{2x\sqrt{x}+8x}{-x+\sqrt{x}+6}$
Giải thích các bước giải:
$B=\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{6x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)$
$=\dfrac{2x\sqrt{x}+16\sqrt{x}+12x}{\left(-x+4\right)\left(\sqrt{x}+2\right)}\::\:\dfrac{-x+3\sqrt{x}}{x\sqrt{x}-2x}$
$=\dfrac{\left(2x\sqrt{x}+16\sqrt{x}+12x\right)\left(x\sqrt{x}-2x\right)}{\left(-x+4\right)\left(\sqrt{x}+2\right)\left(-x+3\sqrt{x}\right)}$
$=\dfrac{2\left(4x^2+x^2\sqrt{x}-4x\sqrt{x}-16x\right)}{\left(-x+4\right)\left(-x+\sqrt{x}+6\right)}$
$=-\dfrac{2x\left(\sqrt{x}+4\right)}{-x+\sqrt{x}+6}$
$=-\dfrac{2x\sqrt{x}+8x}{-x+\sqrt{x}+6}$
