Question

\(\dfrac{\sqrt{x-\sqrt{4\left(x-1\right)}}+\sqrt{x+\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}\left(1-\dfrac{1}{x-1}\right)\) (với \(x>1;x\ne2\))

Answer 1

\(....=\dfrac{\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}}{\sqrt{x^2-4x+4}}\left(\dfrac{x-2}{x-1}\right)\)

\(=\dfrac{\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}}{\sqrt{\left(x-2\right)^2}}.\dfrac{x-2}{x-1}\)

\(=\dfrac{\left|\sqrt{x-1}-1\right|+\sqrt{x-1}+1}{\left|x-2\right|}.\dfrac{x-2}{x-1}\)

Kết hợp điều kiện, ta xét các khoảng sau

\(x>2\) thì

\(..=\dfrac{\sqrt{x-1}-1+\sqrt{x-1}+1}{x-2}.\dfrac{x-2}{x-1}\)

\(=\dfrac{2\sqrt{x-1}}{x-1}\)

\(1< x< 2\) thì

\(..=\dfrac{1-\sqrt{x-1}+\sqrt{x-1}+1}{2-x}.\dfrac{x-2}{x-1}\)

\(=\dfrac{-2}{x-1}\)

Vậy.....