Đáp án:
$\begin{array}{l}
1)\\
a)15\\
b)3 - \sqrt 3 \\
c)4 - \sqrt 2 \\
d)\dfrac{1}{2}\\
2)\\
a)26\\
b){\left( {2 - a} \right)^2} + 2a + 2\sqrt a
\end{array}$
Giải thích các bước giải:
$\begin{array}{l}
1)\\
a)\left( {\dfrac{1}{2}\sqrt {48} + \dfrac{2}{3}\sqrt {243} - \dfrac{3}{5}\sqrt {75} } \right)\sqrt 3 \\
= \left( {\dfrac{1}{2}.4\sqrt 3 + \dfrac{2}{3}.9\sqrt 3 - \dfrac{3}{5}.5\sqrt 3 } \right)\sqrt 3 \\
= \left( {2\sqrt 3 + 6\sqrt 3 - 3\sqrt 3 } \right)\sqrt 3 \\
= 5\sqrt 3 .\sqrt 3 \\
= 15\\
b)\sqrt {12 - 6\sqrt 3 } \\
= \sqrt {{{\left( {3 - \sqrt 3 } \right)}^2}} \\
= \left| {3 - \sqrt 3 } \right|\\
= 3 - \sqrt 3 \\
c)\sqrt {18 - 8\sqrt 2 } \\
= \sqrt {{{\left( {4 - \sqrt 2 } \right)}^2}} \\
= \left| {4 - \sqrt 2 } \right|\\
= 4 - \sqrt 2 \\
d)\dfrac{{\sqrt {6 + 2\sqrt 5 } }}{{\sqrt {20} + 2}}\\
= \dfrac{{\sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} }}{{2 + 2\sqrt 5 }}\\
= \dfrac{{\left| {\sqrt 5 + 1} \right|}}{{2\left( {\sqrt 5 + 1} \right)}}\\
= \dfrac{{\sqrt 5 + 1}}{{2\left( {\sqrt 5 + 1} \right)}}\\
= \dfrac{1}{2}\\
2)\\
a)\sqrt {13a} \sqrt {\dfrac{{52}}{a}} \left( {a > 0} \right)\\
= \sqrt {13a.\dfrac{{52}}{a}} \\
= \sqrt {13.52} \\
= \sqrt {13.13.4} \\
= \sqrt {{{\left( {2.13} \right)}^2}} \\
= 2.13\\
= 26\\
b){\left( {2 - a} \right)^2} + 10\sqrt {0,1a} .\sqrt {0,4a} + \sqrt {4{a^2}} \left( {a > 0} \right)\\
= {\left( {2 - a} \right)^2} + 10\sqrt {0,1a.0,4a} + \sqrt {{{\left( {2\sqrt a } \right)}^2}} \\
= {\left( {2 - a} \right)^2} + 10.\sqrt {{{\left( {a.0,2} \right)}^2}} + 2\sqrt a \\
= {\left( {2 - a} \right)^2} + 10.a.0,2 + 2\sqrt a \\
= {\left( {2 - a} \right)^2} + 2a + 2\sqrt a
\end{array}$
