Đáp án:
\(\begin{array}{l}
1,\\
I = 3 + 5\sqrt 5 \\
2,\\
J = 2 + 2\sqrt 2 \\
3,\\
K = \dfrac{1}{2}\\
4,\\
L = \sqrt 5 + 2
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
I = \left( {\sqrt {54} + \sqrt {45} } \right)\left( {\sqrt 6 - \sqrt 5 } \right) + \sqrt {245} - \sqrt {20} \\
= \left( {\sqrt {9.6} + \sqrt {9.5} } \right)\left( {\sqrt 6 - \sqrt 5 } \right) + \sqrt {49.5} - \sqrt {4.5} \\
= \left( {\sqrt 9 .\sqrt 6 + \sqrt 9 .\sqrt 5 } \right)\left( {\sqrt 6 - \sqrt 5 } \right) + \sqrt {{7^2}.5} - \sqrt {{2^2}.5} \\
= \left( {3\sqrt 6 + 3\sqrt 5 } \right)\left( {\sqrt 6 - \sqrt 5 } \right) + 7\sqrt 5 - 2\sqrt 5 \\
= 3.\left( {\sqrt 6 + \sqrt 5 } \right)\left( {\sqrt 6 - \sqrt 5 } \right) + 5\sqrt 5 \\
= 3.\left( {{{\sqrt 6 }^2} - {{\sqrt 5 }^2}} \right) + 5\sqrt 5 \\
= 3.\left( {6 - 5} \right) + 5\sqrt 5 \\
= 3.1 + 5\sqrt 5 \\
= 3 + 5\sqrt 5 \\
2,\\
J = \left( {\dfrac{{3 + 2\sqrt 3 }}{{\sqrt 3 }} + \dfrac{{2 + \sqrt 2 }}{{\sqrt 2 + 1}}} \right) - \dfrac{1}{{\sqrt 3 + \sqrt 2 }}\\
= \left( {\dfrac{{{{\sqrt 3 }^2} + 2\sqrt 3 }}{{\sqrt 3 }} + \dfrac{{{{\sqrt 2 }^2} + \sqrt 2 }}{{\sqrt 2 + 1}}} \right) - \dfrac{{\sqrt 3 - \sqrt 2 }}{{\left( {\sqrt 3 + \sqrt 2 } \right)\left( {\sqrt 3 - \sqrt 2 } \right)}}\\
= \left( {\dfrac{{\sqrt 3 \left( {\sqrt 3 + 2} \right)}}{{\sqrt 3 }} + \dfrac{{\sqrt 2 .\left( {\sqrt 2 + 1} \right)}}{{\sqrt 2 + 1}}} \right) - \dfrac{{\sqrt 3 - \sqrt 2 }}{{{{\sqrt 3 }^2} - {{\sqrt 2 }^2}}}\\
= \left[ {\left( {\sqrt 3 + 2} \right) + \sqrt 2 } \right] - \dfrac{{\sqrt 3 - \sqrt 2 }}{{3 - 2}}\\
= \sqrt 3 + 2 + \sqrt 2 - \dfrac{{\sqrt 3 - \sqrt 2 }}{1}\\
= \sqrt 3 + 2 + \sqrt 2 - \sqrt 3 + \sqrt 2 \\
= 2 + 2\sqrt 2 \\
3,\\
K = \left( {\dfrac{1}{{3 - \sqrt 5 }} - \dfrac{1}{{3 + \sqrt 5 }}} \right):\dfrac{{5 - \sqrt 5 }}{{\sqrt 5 - 1}}\\
= \dfrac{{\left( {3 + \sqrt 5 } \right) - \left( {3 - \sqrt 5 } \right)}}{{\left( {3 - \sqrt 5 } \right)\left( {3 + \sqrt 5 } \right)}}:\dfrac{{{{\sqrt 5 }^2} - \sqrt 5 }}{{\sqrt 5 - 1}}\\
= \dfrac{{3 + \sqrt 5 - 3 + \sqrt 5 }}{{{3^2} - {{\sqrt 5 }^2}}}:\dfrac{{\sqrt 5 .\left( {\sqrt 5 - 1} \right)}}{{\sqrt 5 - 1}}\\
= \dfrac{{2\sqrt 5 }}{{9 - 5}}:\sqrt 5 \\
= \dfrac{{2\sqrt 5 }}{4}:\sqrt 5 \\
= \dfrac{2}{4} = \dfrac{1}{2}\\
4,\\
L = \left( {2 + \dfrac{{3 - \sqrt 3 }}{{1 - \sqrt 3 }}} \right)\left( {2 + \dfrac{{3 + \sqrt 3 }}{{1 + \sqrt 3 }}} \right):\left( {\sqrt 5 - 2} \right)\\
= \left( {2 + \dfrac{{{{\sqrt 3 }^2} - \sqrt 3 }}{{1 - \sqrt 3 }}} \right)\left( {2 + \dfrac{{{{\sqrt 3 }^2} + \sqrt 3 }}{{1 + \sqrt 3 }}} \right):\left( {\sqrt 5 - 2} \right)\\
= \left( {2 + \dfrac{{\sqrt 3 .\left( {\sqrt 3 - 1} \right)}}{{1 - \sqrt 3 }}} \right)\left( {2 + \dfrac{{\sqrt 3 .\left( {\sqrt 3 + 1} \right)}}{{1 + \sqrt 3 }}} \right):\left( {\sqrt 5 - 2} \right)\\
= \left( {2 - \sqrt 3 } \right).\left( {2 + \sqrt 3 } \right):\left( {\sqrt 5 - 2} \right)\\
= \left( {{2^2} - {{\sqrt 3 }^2}} \right):\left( {\sqrt 5 - 2} \right)\\
= \left( {4 - 3} \right):\left( {\sqrt 5 - 2} \right)\\
= 1:\left( {\sqrt 5 - 2} \right)\\
= \dfrac{1}{{\sqrt 5 - 2}}\\
= \dfrac{{\sqrt 5 + 2}}{{\left( {\sqrt 5 - 2} \right).\left( {\sqrt 5 + 2} \right)}}\\
= \dfrac{{\sqrt 5 + 2}}{{{{\sqrt 5 }^2} - {2^2}}}\\
= \dfrac{{\sqrt 5 + 2}}{{5 - 4}}\\
= \dfrac{{\sqrt 5 + 2}}{1}\\
= \sqrt 5 + 2
\end{array}\)
