Đáp án: $B=(x-2\sqrt{x}-1)\cdot(\sqrt{x}-1)$
Giải thích các bước giải:
Ta có:
$B=(x-\sqrt{x}-\dfrac{x\sqrt{x}+1}{x-\sqrt{x}+1}):\dfrac{\sqrt{x}+1}{x-1}$
$\to B=(x-\sqrt{x}-\dfrac{(\sqrt{x})^3+1}{x-\sqrt{x}+1}):\dfrac{\sqrt{x}+1}{(\sqrt{x}+1)(\sqrt{x}-1)}$
$\to B=(x-\sqrt{x}-\dfrac{(\sqrt{x}+1)(x-\sqrt{x}+1)}{x-\sqrt{x}+1}):\dfrac{\sqrt{x}+1}{(\sqrt{x}+1)(\sqrt{x}-1)}$
$\to B=(x-\sqrt{x}-(\sqrt{x}+1)):\dfrac{1}{\sqrt{x}-1}$
$\to B=(x-2\sqrt{x}-1)\cdot(\sqrt{x}-1)$