Đáp án:
\(\begin{array}{l}
a,\,\,\,15\sqrt 2 \\
b,\,\,\,1\\
c,\,\,\, - 3
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
2\sqrt {18} - \dfrac{3}{4}\sqrt {32} + 3\sqrt {50} - 2\sqrt {1\dfrac{7}{2}} \\
= 2\sqrt {9.2} - \dfrac{3}{4}\sqrt {16.2} + 3\sqrt {25.2} - 2\sqrt {\dfrac{9}{2}} \\
= 2\sqrt {{3^2}.2} - \dfrac{3}{4}\sqrt {{4^2}.2} + 3\sqrt {{5^2}.2} - 2\sqrt {\dfrac{9}{4}.2} \\
= 2.3\sqrt 2 - \dfrac{3}{4}.4\sqrt 2 + 3.5\sqrt 2 - 2.\sqrt {{{\left( {\dfrac{3}{2}} \right)}^2}.2} \\
= 6\sqrt 2 - 3\sqrt 2 + 15\sqrt 2 - 2.\dfrac{3}{2}\sqrt 2 \\
= 6\sqrt 2 - 3\sqrt 2 + 15\sqrt 2 - 3\sqrt 2 \\
= 15\sqrt 2 \\
b,\\
\sqrt {17 + 3\sqrt {32} } \sqrt {{{\left( {\sqrt 8 - 3} \right)}^2}} \\
= \sqrt {17 + 3.\sqrt {{4^2}.2} } .\left| {\sqrt 8 - 3} \right|\\
= \sqrt {17 + 3.4\sqrt 2 } .\left( {3 - \sqrt 8 } \right)\\
= \sqrt {9 + 12\sqrt 2 + 8} .\left( {3 - \sqrt {{2^2}.2} } \right)\\
= \sqrt {{3^2} + 2.3.2\sqrt 2 + {{\left( {2\sqrt 2 } \right)}^2}} .\left( {3 - 2\sqrt 2 } \right)\\
= \sqrt {{{\left( {3 + 2\sqrt 2 } \right)}^2}} .\left( {3 - 2\sqrt 2 } \right)\\
= \left( {3 + 2\sqrt 2 } \right).\left( {3 - 2\sqrt 2 } \right)\\
= {3^2} - {\left( {2\sqrt 2 } \right)^2}\\
= 9 - 8\\
= 1\\
c,\\
\dfrac{1}{{\sqrt {10} + 3}} + \dfrac{{5\sqrt 2 - 2\sqrt 5 }}{{\sqrt 2 - \sqrt 5 }}\\
= \dfrac{{\sqrt {10} - 3}}{{\left( {\sqrt {10} + 3} \right)\left( {\sqrt {10} - 3} \right)}} + \dfrac{{{{\sqrt 5 }^2}.\sqrt 2 - {{\sqrt 2 }^2}.\sqrt 5 }}{{\sqrt 2 - \sqrt 5 }}\\
= \dfrac{{\sqrt {10} - 3}}{{{{\sqrt {10} }^2} - {3^2}}} + \dfrac{{\sqrt 5 .\sqrt 2 .\left( {\sqrt 5 - \sqrt 2 } \right)}}{{\sqrt 2 - \sqrt 5 }}\\
= \dfrac{{\sqrt {10} - 3}}{{10 - 9}} - \sqrt 5 .\sqrt 2 \\
= \dfrac{{\sqrt {10} - 3}}{1} - \sqrt {10} \\
= \sqrt {10} - 3 - \sqrt {10} \\
= - 3
\end{array}\)
