Giải thích các bước giải:
Đặt $x-\sqrt{x^2-1}=t$
$\to t=\dfrac{x^2-(x^2-1)}{x+\sqrt{x^2-1}}=\dfrac{1}{x+\sqrt{x^2-1}}$
$\to x+\sqrt{x^2-1}=\dfrac{1}{t}$
$\to \sqrt[4]{t}+\sqrt{\dfrac{1}{t}}=2$
Đặt $\sqrt[4]{t}=y\to \sqrt{t}=y^2,y\ge 0$
$\to y+\dfrac{1}{y^2}=2$
$\to y^3-2y^2+1=0$
$\to \left(y-1\right)\left(y^2-y-1\right)=0$
$\to y=1, y=\dfrac{1+\sqrt{5}}{2}$
+) $y=1\to t=y^4=1\to x-\sqrt{x^2-1}=1$
$\to x-1=\sqrt{x^2-1}$
$\to (x-1)^2=x^2-1$
$\to x^2-2x+1=x^2-1$
$\to x=1$
+) $y=\dfrac{1+\sqrt{5}}{2}$
$\to x-\sqrt{x^2-1}=(\dfrac{1+\sqrt{5}}{2})^4$
$\to x+\sqrt{x^2-1}=\dfrac{1}{(\dfrac{1+\sqrt{5}}{2})^4}$
$\to x-\sqrt{x^2-1}+ x+\sqrt{x^2-1}=(\dfrac{1+\sqrt{5}}{2})^4+\dfrac{1}{(\dfrac{1+\sqrt{5}}{2})^4}$
$\to x=\dfrac{1}{2}((\dfrac{1+\sqrt{5}}{2})^4+\dfrac{1}{(\dfrac{1+\sqrt{5}}{2})^4})$
