Đáp án:
\(\begin{array}{l}
B1:\\
a)2\sqrt 5 \\
b)\sqrt {10} - \sqrt 5 - \sqrt {30} \\
c)1\\
d)\sqrt 5 \\
B2:\\
a)x = 3\\
b)x = 0
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)2.3\sqrt 5 - 3.6\sqrt 2 - \dfrac{1}{2}.8\sqrt 5 + 6.3\sqrt 2 \\
= 6\sqrt 5 - 4\sqrt 5 - 18\sqrt 2 + 18\sqrt 2 \\
= 2\sqrt 5 \\
b)\left| {\sqrt 5 - \sqrt {10} } \right| - \sqrt {10\left( {2 + 1} \right)} \\
= \sqrt {10} - \sqrt 5 - \sqrt {30} \\
c){\left( {\sqrt 2 + \sqrt 3 } \right)^2}.\sqrt {25 - 2.5.2\sqrt 6 + 24} \\
= \left( {5 + 2\sqrt 6 } \right).\sqrt {{{\left( {5 - 2\sqrt 6 } \right)}^2}} \\
= \left( {5 + 2\sqrt 6 } \right)\left( {5 - 2\sqrt 6 } \right)\\
= 25 - 24 = 1\\
d)\dfrac{2}{{\sqrt {8 - 2\sqrt {15} } }} - \sqrt {\dfrac{{3\sqrt 2 + 3\sqrt 3 }}{{\sqrt 3 + \sqrt 2 }}} \\
= \dfrac{2}{{\sqrt {5 - 2.\sqrt 5 .\sqrt 3 + 3} }} - \sqrt {\dfrac{{3\left( {\sqrt 3 + \sqrt 2 } \right)}}{{\sqrt 3 + \sqrt 2 }}} \\
= \dfrac{2}{{\sqrt {{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}} }} - \sqrt 3 \\
= \dfrac{2}{{\sqrt 5 - \sqrt 3 }} - \sqrt 3 \\
= \dfrac{{2 - \sqrt {15} + 3}}{{\sqrt 5 - \sqrt 3 }}\\
= \dfrac{{5 - \sqrt {15} }}{{\sqrt 5 - \sqrt 3 }} = \dfrac{{\sqrt 5 \left( {\sqrt 5 - \sqrt 3 } \right)}}{{\sqrt 5 - \sqrt 3 }}\\
= \sqrt 5 \\
B2:\\
a)DK:x \ge \dfrac{1}{2}\\
\sqrt {8\left( {2x - 1} \right)} + \sqrt {18\left( {2x - 1} \right)} - \sqrt {32\left( {2x - 1} \right)} = \sqrt {10} \\
\to \left( {2\sqrt 2 + 3\sqrt 2 - 4\sqrt 2 } \right)\sqrt {2x - 1} = \sqrt {10} \\
\to \sqrt 2 .\sqrt {2x - 1} = \sqrt {10} \\
\to \sqrt {2x - 1} = \sqrt 5 \\
\to 2x - 1 = 5\\
\to x = 3\\
b)\sqrt {{{\left( {x - 3} \right)}^2}} = x + 3\\
\to \left| {x - 3} \right| = x + 3\\
\to \left[ \begin{array}{l}
x - 3 = x + 3\\
x - 3 = - x - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
- 3 = 3\left( l \right)\\
2x = 0
\end{array} \right.\\
\to x = 0
\end{array}\)
