a) Tính:  \(\sqrt {{{\left( {2 -...

C

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

\(\begin{array}{l}
\sqrt {{{\left( {2 - \sqrt 5 } \right)}^2}} - \sqrt {\dfrac{8}{{7 - 3\sqrt 5 }}} \\
= \left| {2 - \sqrt 5 } \right| - \sqrt {\dfrac{{16}}{{14 - 6\sqrt 5 }}} \\
= \left| {2 - \sqrt 5 } \right| - \dfrac{{\sqrt {16} }}{{\sqrt {14 - 6\sqrt 5 } }}\\
= - \left( {2 - \sqrt 5 } \right) - \dfrac{4}{{\sqrt {{3^2} - 2.3.\sqrt 5 + {{\sqrt 5 }^2}} }}\left( {Do{\rm{ }}2 - \sqrt 5 < 0} \right)\\
= - 2 + \sqrt 5 - \dfrac{4}{{\sqrt {{{\left( {3 - \sqrt 5 } \right)}^2}} }}\\
= - 2 + \sqrt 5 - \dfrac{4}{{3 - \sqrt 5 }}\\
= - 2 + \sqrt 5 - \dfrac{{4.\left( {3 + \sqrt 5 } \right)}}{{\left( {3 - \sqrt 5 } \right)\left( {3 + \sqrt 5 } \right)}}\\
= - 2 + \sqrt 5 - \dfrac{{4.\left( {3 + \sqrt 5 } \right)}}{4}\\
= - 2 + \sqrt 5 - 3 - \sqrt 5 = - 5
\end{array}\)

b)

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