Đáp án:
\(\begin{array}{l}
c)\dfrac{{\sqrt 5 }}{2}\\
e)\dfrac{{4 - 5\sqrt 6 }}{{10}}\\
b)6\\
d)3\\
f)\sqrt 3 + 1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
c)\dfrac{{\sqrt 5 \left( {\sqrt 2 - \sqrt 3 } \right)}}{{2\left( {\sqrt 2 - \sqrt 3 } \right)}} = \dfrac{{\sqrt 5 }}{2}\\
e)\dfrac{{2.2\sqrt 2 - 2\sqrt 3 }}{{3\sqrt 2 - 4\sqrt 3 }} - \dfrac{{\sqrt 5 + 3\sqrt 3 }}{{\sqrt {30} + 9\sqrt 2 }}\\
= \dfrac{{4\sqrt 2 - 2\sqrt 3 }}{{3\sqrt 2 - 4\sqrt 3 }} - \dfrac{{\sqrt 5 + 3\sqrt 3 }}{{\sqrt 6 \left( {\sqrt 5 + 3\sqrt 3 } \right)}}\\
= \dfrac{{\left( {4\sqrt 2 - 2\sqrt 3 } \right)\left( {3\sqrt 2 + 4\sqrt 3 } \right)}}{{18 - 48}} - \dfrac{1}{{\sqrt 6 }}\\
= \dfrac{{12.2 + 16\sqrt 6 - 6\sqrt 6 - 12.3}}{{ - 30}} - \dfrac{1}{{\sqrt 6 }}\\
= \dfrac{{ - 12 + 10\sqrt 6 }}{{ - 30}} - \dfrac{1}{{\sqrt 6 }}\\
= \dfrac{{6 - 5\sqrt 6 }}{{15}} - \dfrac{1}{{\sqrt 6 }}\\
= \dfrac{{6\sqrt 6 - 30 - 15}}{{15\sqrt 6 }}\\
= \dfrac{{6\sqrt 6 - 45}}{{15\sqrt 6 }} = \dfrac{{4 - 5\sqrt 6 }}{{10}}\\
b)\sqrt {17 - 3.4\sqrt 2 } + \sqrt {17 + 3.4\sqrt 2 } \\
= \sqrt {9 - 2.3.2\sqrt 2 + 8} + \sqrt {9 + 2.3.2\sqrt 2 + 8} \\
= \sqrt {{{\left( {3 - 2\sqrt 2 } \right)}^2}} + \sqrt {{{\left( {3 + 2\sqrt 2 } \right)}^2}} \\
= 3 - 2\sqrt 2 + 3 + 2\sqrt 2 = 6\\
d)\left( {\sqrt {3 - 2.\sqrt 3 .\sqrt 2 + 2} + \sqrt 2 } \right)\sqrt 3 \\
= \left( {\sqrt {{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}} + \sqrt 2 } \right)\sqrt 3 \\
= \left( {\sqrt 3 - \sqrt 2 + \sqrt 2 } \right)\sqrt 3 = 3\\
f)\sqrt {\sqrt 2 + 2\sqrt 3 + \sqrt {16 - 2.4.\sqrt 2 + 2} } \\
= \sqrt {\sqrt 2 + 2\sqrt 3 + \sqrt {{{\left( {4 - \sqrt 2 } \right)}^2}} } \\
= \sqrt {\sqrt 2 + 2\sqrt 3 + 4 - \sqrt 2 } \\
= \sqrt {3 + 2.\sqrt 3 .1 + 1} \\
= \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} \\
= \sqrt 3 + 1
\end{array}\)
