Đáp án:
\(\begin{array}{l}
1A,\\
a,\\
3\sqrt 3 x\\
b,\\
- 2\sqrt {2x} y\\
1B,\\
a,\\
5x\sqrt x \\
b,\\
4\sqrt 3 \sqrt x {y^2}\\
2A,\\
a,\\
\sqrt {13{a^2}} \\
b,\\
- \sqrt { - 15a} \\
2B,\\
a,\\
\sqrt {3a} \\
b,\\
- \sqrt {2{a^2}}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1A,\\
a,\\
x \ge 0 \Rightarrow \left| x \right| = x\\
\sqrt {27{x^2}} = \sqrt {9.3{x^2}} = \sqrt {{3^2}.{{\sqrt 3 }^2}.{x^2}} = \sqrt {{{\left( {3\sqrt 3 x} \right)}^2}} \\
= \left| {3\sqrt 3 x} \right| = 3\sqrt 3 \left| x \right| = 3\sqrt 3 x\\
b,\\
x \ge 0,y \le 0 \Rightarrow \left\{ \begin{array}{l}
\sqrt x \,\,có\,\,nghĩa\\
\left| y \right| = - y
\end{array} \right.\\
\sqrt {8x{y^2}} = \sqrt {4.2.x.{y^2}} = \sqrt {{2^2}.{{\sqrt 2 }^2}.{{\sqrt x }^2}.{y^2}} \\
= \sqrt {{{\left( {2.\sqrt 2 .\sqrt x .y} \right)}^2}} = \sqrt {{{\left( {2y\sqrt {2x} } \right)}^2}} \\
= \left| {2y\sqrt {2x} } \right| = 2\sqrt {2x} .\left| y \right| = 2\sqrt {2x} .\left( { - y} \right) = - 2\sqrt {2x} y\\
1B,\\
a,\\
x > 0 \Rightarrow \left\{ \begin{array}{l}
\sqrt x \,\,có\,\,nghĩa\\
\left| x \right| = x
\end{array} \right.\\
\sqrt {25{x^3}} = \sqrt {25.{x^2}.x} = \sqrt {{5^2}.{x^2}.{{\sqrt x }^2}} = \sqrt {{{\left( {5.x.\sqrt x } \right)}^2}} \\
= \left| {5x\sqrt x } \right| = 5.\left| x \right|.\left| {\sqrt x } \right| = 5x\sqrt x \\
b,\\
x \ge 0 \Rightarrow \sqrt x \,\,có\,\,nghĩa\\
{y^2} \ge 0,\,\,\forall y \Rightarrow \left| {{y^2}} \right| = {y^2}\\
\sqrt {48x{y^4}} = \sqrt {16.3.x.{y^4}} = \sqrt {{4^2}.{{\sqrt 3 }^2}.{{\sqrt x }^2}.{{\left( {{y^2}} \right)}^2}} \\
= \sqrt {{{\left( {4.\sqrt 3 .\sqrt x .{y^2}} \right)}^2}} = \left| {4\sqrt 3 .\sqrt x .{y^2}} \right|\\
= 4\sqrt 3 .\left| {\sqrt x } \right|.\left| {{y^2}} \right| = 4\sqrt 3 \sqrt x {y^2}\\
2A,\\
a,\\
a \ge 0 \Rightarrow \sqrt {{a^2}} = \left| a \right| = a\\
a\sqrt {13} = \sqrt {{a^2}} .\sqrt {13} = \sqrt {{a^2}.13} = \sqrt {13{a^2}} \\
b,\\
a < 0 \Rightarrow \sqrt {{a^2}} = \left| a \right| = - a\\
a.\sqrt {\dfrac{{ - 15}}{a}} = - \sqrt {{a^2}} .\sqrt {\dfrac{{ - 15}}{a}} = - \sqrt {{a^2}.\dfrac{{ - 15}}{a}} = - \sqrt { - 15a} \\
2B,\\
a,\\
a > 0 \Rightarrow \sqrt {{a^2}} = \left| a \right| = a\\
\dfrac{a}{2}\sqrt {\dfrac{{12}}{a}} = \dfrac{{\sqrt {{a^2}} }}{{\sqrt 4 }}.\sqrt {\dfrac{{12}}{a}} = \sqrt {\dfrac{{{a^2}.12}}{{4.a}}} = \sqrt {3a} \\
b,\\
a \le 0 \Rightarrow \sqrt {{a^2}} = \left| a \right| = - a\\
a\sqrt 2 = - \sqrt {{a^2}} .\sqrt 2 = - \sqrt {{a^2}.2} = - \sqrt {2{a^2}}
\end{array}\)
