Cho biểu thức \(A = \dfrac{{\sqrt {x - \sqrt {4\left( {x - 1} \right)} }  + \sqrt {x + \sqrt {4\left( {x - 1} \right)} } }}{{\sqrt {{x^2} - 4\left( {x - 1} \right)} }}.\left( {1 - \dfrac{1}{{x - 1}}} \right),\)trong đó \(x > 1,x \ne 2\).

Rút gọn biểu thức \(A\).

Điều kiện: \(x > 1,\,\,x \ne 2.\)

\(\begin{array}{l}A = \dfrac{{\sqrt {x - 4\left( {x - 1} \right)}  + \sqrt {x + \sqrt {4\left( {x - 1} \right)} } }}{{\sqrt {{x^2} - 4x + 4} }}.\left( {1 - \dfrac{1}{{x - 1}}} \right)\\\,\,\,\,\, = \dfrac{{\sqrt {x - 1 - 2\sqrt {x - 1}  + 1}  + \sqrt {x - 1 + 2\sqrt {x - 1}  + 1} }}{{\sqrt {{{\left( {x - 2} \right)}^2}} }}.\left( {\dfrac{{x - 2}}{{x - 1}}} \right)\\\,\,\,\, = \dfrac{{\sqrt {{{\left( {\sqrt {x - 1}  - 1} \right)}^2}}  + \sqrt {{{\left( {\sqrt {x - 1}  + 1} \right)}^2}} }}{{\sqrt {{{\left( {x - 2} \right)}^2}} }}.\dfrac{{x - 2}}{{x - 1}}\\\,\,\,\, = \dfrac{{\left| {\sqrt {x - 1}  - 1} \right| + \left| {\sqrt {x - 1}  + 1} \right|}}{{\left| {x - 2} \right|}}.\dfrac{{x - 2}}{{x - 1}}\end{array}\)

+) Nếu \(1 < x < 2\) thì \(\left| {\sqrt {x - 1}  - 1} \right| = 1 - \sqrt {x - 1} \)

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

+) Nếu \(x > 2\) thì \(\left| {\sqrt {x - 1}  - 1} \right| = \sqrt {x - 1}  - 1\)

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

Vậy \(A = \left\{ \begin{array}{l}\dfrac{{ - 2}}{{x - 1}}\,\,\,\,khi\,\,\,1 < x < 2\\\dfrac{2}{{\sqrt {x - 1} }}\,\,\,\,khi\,\,\,x > 2\end{array} \right..\)