Bài 286: $x+\frac{x}{\sqrt{x^2-1}}=2\sqrt{2}$ Nghiệm là $x=\sqrt{2}$
ĐK: $x^2 \geq 1$
Ta có:
$x(1+\dfrac{1}{\sqrt{x^2-1}})=2\sqrt{2}$
$\rightarrow x >0$
$\iff x-\sqrt{2}+\dfrac{x}{\sqrt{x^2-1}}-\sqrt{2}=0$
$\iff \dfrac{x^2-2}{x+\sqrt{2}}-\dfrac{x^2-2}{(x+\sqrt{2(x^2-1)})\sqrt{x^2-1}}=0$
$\iff x^2-2=0$ v $\dfrac{1}{x+\sqrt{2}}-\dfrac{1}{(x+\sqrt{2(x^2-1)})\sqrt{x^2-1}}=0$
(2) $\iff x+\sqrt{2}=x\sqrt{x^2-1}+(x^2-1)\sqrt{2}$
$\iff x(\sqrt{x^2-1}-1)+\sqrt{2}(x^2-2)=0$
$\iff \dfrac{x(x^2-2)}{x^2+2}+(x^2-2)\sqrt{2}=0$
$\iff (x^2-2)(\dfrac{x}{x^2+2}+\sqrt{2})=0$
$\iff x^2-2=0$
$\iff x=\sqrt{2}$ ($x >0$)