Đáp án: `a)P=\frac{2}{x-1}`
$b)x=2$
Giải thích các bước giải:
$ĐKXĐ:x>0;x\neq1$
`a)P=(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}).\frac{1}{2x\sqrt{x}}`
`=\frac{(\sqrt{x}+1)^2-(\sqrt{x}-1)^2+4\sqrt{x}(\sqrt{x}+1)(\sqrt{x}-1)}{(\sqrt{x}+1)(\sqrt{x}-1)}.\frac{1}{2x\sqrt{x}}`
`=\frac{(x+1+2\sqrt{x})-(x-2\sqrt{x}+1)+4\sqrt{x}(x-1)}{(\sqrt{x}+1)(\sqrt{x}-1)}.\frac{1}{2x\sqrt{x}}`
`=\frac{x+1+2\sqrt{x}-x+2\sqrt{x}-1+4x\sqrt{x}-4\sqrt{x}}{x-1}.\frac{1}{2x\sqrt{x}}`
`=\frac{4x\sqrt{x}}{x-1}.\frac{1}{2x\sqrt{x}}`
`=\frac{2}{x-1}`
$\text{b)Để P=x}$
`⇔\frac{2}{x-1}=x`
$⇒x(x-1)=2$
$⇔x^2-x-2=0$
$⇔x^2+x-2x-2=0$
$⇔x(x+1)-2(x+1)=0$
$⇔(x+1)(x-2)=0$
$⇔\left[ \begin{array}{l}x+1=0\\x-2=0\end{array} \right.$
$⇔\left[ \begin{array}{l}x=-1\\x=2\end{array} \right.$
Kết hợp với ĐK, ta được: $x=2$
