Giải pt: 
1)$2x+\frac{x-1}{x}-\sqrt{1-\frac{1}{x}}-3\sqrt{x-\frac{1}{x}}=0$
2)$1+x-x^2=\sqrt{4x^2-1}-\sqrt{2x}+1$
DKXD:..
Dùng bdt cosi
$1.\sqrt{x-\frac 1x} \le \frac{1+x-\dfrac 1x}{2}$
$\sqrt{1-\frac 1x}=\sqrt{\frac 1x.\left(x-1 \right)}\le\frac{\dfrac 1x+x-1}{2}$
Do đó $\sqrt{x-\frac 1x}+3\sqrt{1-\frac 1x}\le-\frac 1x+2x+1$
$\Leftrightarrow VT\ge0$,Dấu = xảy ra $\Leftrightarrow x=\frac{\sqrt 5+1}{2}$
Nên đó là nghiệm duy nhất của pt
$(2) \Leftrightarrow 4(x^2-x-1)+4\sqrt{4x^2-1}-4\sqrt{2x+1}=0$
$\Leftrightarrow 4x^2-1+4\sqrt{4x^2-1}+1-2(2x+1)-4\sqrt{2x+1}-2=0$
$\Leftrightarrow (2\sqrt{4x^2-1}+1)^2-2(\sqrt{2x+1}+1)^2=0$
$\Leftrightarrow (2\sqrt{4x^2-1}+1)^2-(\sqrt{4x+2}+\sqrt{2})^2=0$

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