Đáp án:
\(\begin{array}{l}
a)6\sqrt 2 \\
- 10\sqrt 3 \\
2\sqrt 2 \\
\left[ \begin{array}{l}
\sqrt b \\
- \sqrt b
\end{array} \right.\\
b)\sqrt {15} - 2\sqrt 3 \\
6\sqrt 2 - 3\sqrt 6 \\
\dfrac{{\sqrt {15} - \sqrt 5 }}{2}\\
c)4 + 2\sqrt 2 \\
\left[ \begin{array}{l}
\left( {1 - x} \right)\sqrt {x - 1} \\
- \left( {1 - x} \right)\sqrt {x - 1}
\end{array} \right.\\
\left( {x\sqrt 3 - x} \right)\sqrt {x - x\sqrt 3 } \\
d)5\left( {{a^2} + 10a + 25} \right)\sqrt {2a + 10} \\
- \left( {{x^3} - 6{x^2} + 9x - 4} \right)\sqrt { - {x^2} + 5x - 4}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)3.2\sqrt 2 = 6\sqrt 2 \\
- 2.5\sqrt 3 = - 10\sqrt 3 \\
\dfrac{2}{5}.5\sqrt 2 = 2\sqrt 2 \\
\dfrac{1}{a}.\left| a \right|\sqrt b = \left[ \begin{array}{l}
\dfrac{1}{a}.a\sqrt b \\
- \dfrac{1}{a}.a\sqrt b
\end{array} \right. = \left[ \begin{array}{l}
\sqrt b \\
- \sqrt b
\end{array} \right.\\
b)\sqrt {3{{\left( {2 - \sqrt 5 } \right)}^2}} \\
= \left| {2 - \sqrt 5 } \right|\sqrt 3 \\
= \left( {\sqrt 5 - 2} \right)\sqrt 3 \left( {do:\sqrt 5 > 2} \right)\\
= \sqrt {15} - 2\sqrt 3 \\
\sqrt {18{{\left( {2 - \sqrt 3 } \right)}^2}} = \left| {2 - \sqrt 3 } \right|.3\sqrt 2 \\
= \left( {2 - \sqrt 3 } \right).3\sqrt 2 \\
= 6\sqrt 2 - 3\sqrt 6 \\
\dfrac{{\sqrt {5{{\left( {1 - \sqrt 3 } \right)}^2}} }}{2} = \dfrac{{\left| {1 - \sqrt 3 } \right|\sqrt 5 }}{2}\\
= \dfrac{{\left( {\sqrt 3 - 1} \right)\sqrt 5 }}{2} = \dfrac{{\sqrt {15} - \sqrt 5 }}{2}\\
c)\dfrac{{2\sqrt 2 }}{{\left| {1 - \sqrt 2 } \right|}} = \dfrac{{2\sqrt 2 }}{{\sqrt 2 - 1}} = \dfrac{{2\sqrt 2 \left( {\sqrt 2 + 1} \right)}}{{2 - 1}}\\
= 4 + 2\sqrt 2 \\
\left| {1 - x} \right|\sqrt {1 - x} = \left[ \begin{array}{l}
\left( {1 - x} \right)\sqrt {x - 1} \\
- \left( {1 - x} \right)\sqrt {x - 1}
\end{array} \right.\\
\left| {x\left| {1 - \sqrt 3 } \right|} \right|\sqrt {x\left( {1 - \sqrt 3 } \right)} \\
= \left( {\sqrt 3 - 1} \right)\left| x \right|\sqrt {x - x\sqrt 3 } \\
= \left( {x\sqrt 3 - x} \right)\sqrt {x - x\sqrt 3 } \\
d)5{\left( {a + 5} \right)^2}\sqrt {2\left( {a + 5} \right)} \\
= 5\left( {{a^2} + 10a + 25} \right)\sqrt {2a + 10} \\
\left| {x - 4} \right|{\left( {1 - x} \right)^2}\sqrt {\left( {x - 4} \right)\left( {1 - x} \right)} \\
= - \left( {x - 4} \right){\left( {1 - x} \right)^2}\sqrt {\left( {x - 4} \right)\left( {1 - x} \right)} \\
= - \left( {x - 4} \right)\left( {1 - 2x + {x^2}} \right)\sqrt { - {x^2} + 5x - 4} \\
= - \left( {{x^3} - 6{x^2} + 9x - 4} \right)\sqrt { - {x^2} + 5x - 4}
\end{array}\)