`(√x) / (√x-1) + 3/(√x+1) - (6√x -4) / (x-1)` tìm x để bthuc trên có nghĩa và rút gọn biểu thức trên

ĐKXĐ: $\begin{cases} x\ge0\\x\ne1 \end{cases}$
$\dfrac{\sqrt{x}}{\sqrt{x}-1}$$+$$\dfrac{3}{\sqrt{x}+1}$$-$$\dfrac{6\sqrt{x}-4}{x-1}$

$=$$\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{(\sqrt{x}-1)(\sqrt{x}+1)}$

$=$$\dfrac{x-2\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}$

$=$$\dfrac{(\sqrt{x}-1)^2}{(\sqrt{x}-1)(\sqrt{x}+1)}$

$=$$\dfrac{\sqrt{x}-1}{\sqrt{x}+1}$

@Nobitao@

`#huy`

`(\sqrt{x})/(\sqrt{x}-1)+3/(\sqrt{x}+1)-(6\sqrt{x}-4)/(x-1)`

`=(\sqrt{x}(\sqrt{x}+1)+3(\sqrt{x}-1)-6\sqrt{x}+4)/((\sqrt{x}-1)(\sqrt{x}+1))`

`=(x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4)/((\sqrt{x}-1)(\sqrt{x}+1))`

`=(x-2\sqrt{x}+1)/((\sqrt{x}-1)(\sqrt{x}+1))`

`=((\sqrt{x}-1)^2)/((\sqrt{x}-1)(\sqrt{x}+1))`

`=(\sqrt{x}-1)/(\sqrt{x}+1)`