a) Ta có: \(4 > 3 \Rightarrow \sqrt 4 > \sqrt 3 \Rightarrow 2 > \sqrt 3 > 0\)
Suy ra: \(\sqrt {2 + \sqrt 3 } + \sqrt {2 - \sqrt 3 } > 0\)
Ta có:
\({\left( {\sqrt {2 + \sqrt 3 } + \sqrt {2 - \sqrt 3 } } \right)^2}\)\( = 2 + \sqrt 3 + 2\sqrt {2 + \sqrt 3 } .\sqrt {2 - \sqrt 3 } + 2 - \sqrt 3 \)
\( = 4 + 2\sqrt {4 - 3} = 4 + 2\sqrt 1 = 4 + 2 = 6\)
\({\left( {\sqrt 6 } \right)^2} = 6\)
Vì \({\left( {\sqrt {2 + \sqrt 3 } + \sqrt {2 - \sqrt 3 } } \right)^2} = {\left( {\sqrt 6 } \right)^2}\) nên \(\sqrt {2 + \sqrt 3 } + \sqrt {2 - \sqrt 3 } = \sqrt 6 \)
b) Ta có:
\(\begin{array}{l}\sqrt {\dfrac{4}{{{{\left( {2 - \sqrt 5 } \right)}^2}}}} - \sqrt {\dfrac{4}{{{{\left( {2 + \sqrt 5 } \right)}^2}}}} \\ = \dfrac{{\sqrt 4 }}{{\sqrt {{{\left( {2 - \sqrt 5 } \right)}^2}} }} - \dfrac{{\sqrt 4 }}{{\sqrt {{{\left( {2 + \sqrt 5 } \right)}^2}} }}\\ = \dfrac{2}{{\left| {2 - \sqrt 5 } \right|}} - \dfrac{2}{{\left| {2 + \sqrt 5 } \right|}}\end{array}\)
Do \(\sqrt 5 > 2\) nên
\(\begin{array}{l}
\dfrac{2}{{\left| {2 - \sqrt 5 } \right|}} - \dfrac{2}{{\left| {2 + \sqrt 5 } \right|}}\\
= \dfrac{2}{{\sqrt 5 - 2}} - \dfrac{2}{{2 + \sqrt 5 }}\\
= \dfrac{{2(2 + \sqrt 5 ) - 2\left( {\sqrt 5 - 2} \right)}}{{{{\left( {\sqrt 5 } \right)}^2} - {2^2}}}\\
= \dfrac{8}{1} = 8
\end{array}\)
Vế trái bằng vế phải nên đẳng thức được chứng minh.
