Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
x = 2 + \sqrt 3 \\
\Rightarrow A = \dfrac{{\sqrt 3 \left( {x - 1} \right)}}{{\sqrt {{x^2} - x + 1} }}\\
= \dfrac{{\sqrt 3 \left[ {\left( {2 + \sqrt 3 } \right) - 1} \right]}}{{\sqrt {{{\left( {2 + \sqrt 3 } \right)}^2} - \left( {2 + \sqrt 3 } \right) + 1} }}\\
= \dfrac{{\sqrt 3 .\left( {\sqrt 3 + 1} \right)}}{{\sqrt {\left( {7 + 4\sqrt 3 } \right) - \left( {2 + \sqrt 3 } \right) + 1} }}\\
= \dfrac{{\sqrt 3 \left( {\sqrt 3 + 1} \right)}}{{\sqrt {6 + 3\sqrt 3 } }}\\
= \dfrac{{\sqrt 3 \left( {\sqrt 3 + 1} \right)}}{{\sqrt {3.\left( {2 + \sqrt 3 } \right)} }}\\
= \dfrac{{\sqrt 3 .\left( {\sqrt 3 + 1} \right)}}{{\sqrt 3 .\sqrt {2 + \sqrt 3 } }}\\
= \dfrac{{\sqrt 3 + 1}}{{\sqrt {\dfrac{1}{2}.\left( {4 + 2\sqrt 3 } \right)} }}\\
= \dfrac{{\sqrt 3 + 1}}{{\sqrt {\dfrac{1}{2}.{{\left( {\sqrt 3 + 1} \right)}^2}} }}\\
= \dfrac{{\sqrt 3 + 1}}{{\dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 3 + 1} \right)}}\\
= \sqrt 2 \\
2,\\
A = \sqrt[4]{{56 - 24\sqrt 5 }} - \sqrt[4]{{56 + 24\sqrt 5 }}\\
= \sqrt[4]{{36 - 2.6.2\sqrt 5 + 20}} - \sqrt[4]{{36 + 2.6.2\sqrt 5 + 20}}\\
= \sqrt[4]{{{{\left( {6 - 2\sqrt 5 } \right)}^2}}} - \sqrt[4]{{{{\left( {6 + 2\sqrt 5 } \right)}^2}}}\\
= \sqrt {6 - 2\sqrt 5 } - \sqrt {6 + 2\sqrt 5 } \\
= \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} - \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} \\
= \left( {\sqrt 5 - 1} \right) - \left( {\sqrt 5 + 1} \right)\\
= - 2
\end{array}\)
