Đáp án:
\(I = - \sqrt 2 \)
J=1
Giải thích các bước giải:
\(\begin{array}{l}
I = \dfrac{{\sqrt 2 + \sqrt {10} }}{{2 + \sqrt {6 + 2\sqrt 5 } }} + \dfrac{{\sqrt 2 - \sqrt {10} }}{{2 - \sqrt {6 - 2\sqrt 5 } }}\\
= \dfrac{{\sqrt 2 + \sqrt {10} }}{{2 + \sqrt {5 + 2\sqrt 5 .1 + 1} }} + \dfrac{{\sqrt 2 - \sqrt {10} }}{{2 - \sqrt {5 - 2\sqrt 5 .1 + 1} }}\\
= \dfrac{{\sqrt 2 + \sqrt {10} }}{{2 + \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} }} + \dfrac{{\sqrt 2 - \sqrt {10} }}{{2 - \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} }}\\
= \dfrac{{\sqrt 2 + \sqrt {10} }}{{2 + \sqrt 5 + 1}} + \dfrac{{\sqrt 2 - \sqrt {10} }}{{2 - \sqrt 5 + 1}}\\
= \dfrac{{\sqrt 2 + \sqrt {10} }}{{3 + \sqrt 5 }} + \dfrac{{\sqrt 2 - \sqrt {10} }}{{3 - \sqrt 5 }}\\
= \dfrac{{\left( {\sqrt 2 + \sqrt {10} } \right)\left( {3 - \sqrt 5 } \right) + \left( {\sqrt 2 - \sqrt {10} } \right)\left( {3 + \sqrt 5 } \right)}}{{9 - 5}}\\
= \dfrac{{3\sqrt 2 - \sqrt {10} + 3\sqrt {10} - 5\sqrt 2 + 3\sqrt 2 + \sqrt {10} - 3\sqrt {10} - 5\sqrt 2 }}{4}\\
= \dfrac{{ - 4\sqrt 2 }}{4} = - \sqrt 2 \\
J = \dfrac{{2\sqrt {3 + \sqrt {5 - \sqrt {13 + 4\sqrt 3 } } } }}{{\sqrt 6 + \sqrt 2 }}\\
= \dfrac{{2\sqrt {3 + \sqrt {5 - \sqrt {12 + 2.2\sqrt 3 .1 + 1} } } }}{{\sqrt 6 + \sqrt 2 }}\\
= \dfrac{{2\sqrt {3 + \sqrt {5 - \sqrt {{{\left( {2\sqrt 3 + 1} \right)}^2}} } } }}{{\sqrt 6 + \sqrt 2 }}\\
= \dfrac{{2\sqrt {3 + \sqrt {5 - 2\sqrt 3 - 1} } }}{{\sqrt 6 + \sqrt 2 }}\\
= \dfrac{{2\sqrt {3 + \sqrt {4 - 2\sqrt 3 } } }}{{\sqrt 6 + \sqrt 2 }}\\
= \dfrac{{2\sqrt {3 + \sqrt {3 - 2\sqrt 3 .1 + 1} } }}{{\sqrt 6 + \sqrt 2 }}\\
= \dfrac{{2\sqrt {3 + \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} } }}{{\sqrt 6 + \sqrt 2 }}\\
= \dfrac{{2\sqrt {3 + \sqrt 3 - 1} }}{{\sqrt 6 + \sqrt 2 }}\\
= \dfrac{{\sqrt {2.\left( {2 + \sqrt 3 } \right)} }}{{\sqrt 3 + 1}} = \dfrac{{\sqrt {4 + 2\sqrt 3 } }}{{\sqrt 3 + 1}}\\
= \dfrac{{\sqrt {3 + 2\sqrt 3 .1 + 1} }}{{\sqrt 3 + 1}}\\
= \dfrac{{\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} }}{{\sqrt 3 + 1}} = \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 + 1}} = 1
\end{array}\)
