Câu 2
ĐKXĐ : $x \ge 0$
Dễ thấy $x=0$ là 1 nghiệm của pt
Với $x >0$
$pt\Leftrightarrow (\frac{x+1-1}{\sqrt{x+1}-1})[\frac{(2x^2-2x+1)-(x-1)^2}{\sqrt{2x^2-2x+1}-x+1}]=x\sqrt{x}$$\Leftrightarrow \frac{x^3}{(\sqrt{x+1}-1)(\sqrt{2x^2-2x+1}-x+1)}=x\sqrt{x}$
$\Leftrightarrow \frac{x\sqrt{x}}{(\sqrt{x+1}-1)(\sqrt{2x^2-2x+1}-x+1)}=1$
$\Leftrightarrow (\sqrt{x+1}+1)(\frac{\sqrt x}{\sqrt{2x^2-2x+1}-x+1})=1$
$\Leftrightarrow \sqrt x+\sqrt{x^2+x}=\sqrt{2x^2-2x+1}-x+1$
$\Leftrightarrow x+\sqrt x-1=\frac{2x^2-2x+1-x^2-x}{\sqrt{2x^2-2x+1}+\sqrt{x^2+x}}(*)$
$\Leftrightarrow x+\sqrt x-1=\frac{(x - \sqrt x-1)(x+\sqrt x-1)}{\sqrt{2x^2-2x+1}+\sqrt{x^2+x}}$
$\Leftrightarrow [^{x+\sqrt x-1=0}_{\sqrt{2x^2-2x+1}+\sqrt{x^2+x}+\sqrt x+1=x}$
PT thứ 2 dễ dàng c/m $VT >VP$
$\Rightarrow x+\sqrt x-1=0\Rightarrow x=\frac{3-\sqrt 5}2$