Đáp án:
\[A = {\left( {10 + 1} \right)^2} = {11^2} = 121.\]
Giải thích các bước giải:
\[\begin{array}{l}
A = {\left( {x + 1} \right)^2}\\
x = 5{\left( {\sqrt {2 + \sqrt 3 } + \sqrt {3 - \sqrt 5 } - \sqrt {\frac{5}{2}} } \right)^2} + {\left( {\sqrt {2 - \sqrt 3 } + \sqrt {3 + \sqrt 5 } - \sqrt {\frac{3}{2}} } \right)^2}\\
= 5{\left( {\frac{{\sqrt {4 + 2\sqrt 3 } + \sqrt {6 - 2\sqrt 5 } - \sqrt 5 }}{{\sqrt 2 }}} \right)^2} + {\left( {\frac{{\sqrt {4 - 2\sqrt 3 } + \sqrt {6 + 2\sqrt 5 } - \sqrt 3 }}{{\sqrt 2 }}} \right)^2}\\
= 5{\left( {\frac{{\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} - \sqrt 5 }}{{\sqrt 2 }}} \right)^2} + {\left( {\frac{{\sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} - \sqrt 3 }}{{\sqrt 2 }}} \right)^2}\\
= 5{\left( {\frac{{\left| {\sqrt 3 + 1} \right| + \left| {\sqrt 5 - 1} \right| - \sqrt 5 }}{{\sqrt 2 }}} \right)^2} + {\left( {\frac{{\left| {\sqrt 3 - 1} \right| + \left| {\sqrt 5 + 1} \right| - \sqrt 3 }}{{\sqrt 2 }}} \right)^2}\\
= 5{\left( {\frac{{\sqrt 3 + 1 + \sqrt 5 - 1 - \sqrt 5 }}{{\sqrt 2 }}} \right)^2} + {\left( {\frac{{\sqrt 3 - 1 + \sqrt 5 + 1 - \sqrt 3 }}{{\sqrt 2 }}} \right)^2}\\
= 5{\left( {\frac{{\sqrt 3 }}{{\sqrt 2 }}} \right)^2} + {\left( {\frac{{\sqrt 5 }}{{\sqrt 2 }}} \right)^2} = 5.\frac{3}{2} + \frac{5}{2} = 10\\
\Rightarrow A = {\left( {10 + 1} \right)^2} = {11^2} = 121.
\end{array}\]
