giai pt
$\sqrt{x-1/x}$ + $\sqrt{x^{2}-x}$ =2
ĐKXĐ:  $-1\le x<0$  hoặc  $x\ge1$
$\sqrt{x-\frac{1}{x}}+\sqrt{x^2-x}=2$
$\Leftrightarrow \sqrt{\frac{x^2-1}{x}}-1+\sqrt{x^2-x}-1=0$
$\Leftrightarrow \frac{\frac{x^2-1}{x}-1}{\sqrt{\frac{x^2-1}{x}}+1}+\frac{x^2-x-1}{\sqrt{x^2-x}+1}=0$
$\Leftrightarrow (x^2-x-1)(\frac{1}{x(\sqrt\frac{x^2-1}{x}+1)}+\frac{1}{\sqrt{x^2-x}+1})=0$
TH1:  $x^2-x-1=0\Rightarrow x=\frac{1\pm \sqrt5}{2}$
TH2:  $A=x\sqrt\frac{x^2-1}{x}+x+\sqrt{x^2-x}+1=0$
Nếu  $x>1$ PT này vô nghiệm
Nếu $-1\le x<0$
$x\sqrt\frac{x^2-1}{x}=-\sqrt{x(x^2-1)}=-\sqrt{x^3-x}$
$A=\sqrt{x^2-x}-\sqrt{x^3-x}+x+1$
Do  $x^2-x>x^3-x\Rightarrow \sqrt{x^2-x}-\sqrt{x^3-x}>0$
$x+1\ge 0\Rightarrow A>0$ với mọi  $x$ trong tập xác định
Vậy  $x=\frac{1\pm\sqrt5}{2}$

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