Đáp án:
14)
$\begin{array}{l}
a)3\sqrt 2 < 2\sqrt 5 < \sqrt {23} < 5\\
b)\sqrt {47} < 4\sqrt 3 < 5\sqrt 2 < 2\sqrt {13}
\end{array}$
15)
$\begin{array}{l}
a)A = \dfrac{{11}}{2}\sqrt x \left( {x > 0} \right)\\
b)B = - y - \dfrac{3}{4}\left( {y \le \dfrac{1}{2}} \right)
\end{array}$
Giải thích các bước giải:
Bài 14:
$\begin{array}{l}
a)\left\{ \begin{array}{l}
2\sqrt 5 = \sqrt {4.5} = \sqrt {20} \\
3\sqrt 2 = \sqrt {9.2} = \sqrt {18} \\
5 = \sqrt {25} \\
\sqrt {23} = \sqrt {23}
\end{array} \right.\\
\Rightarrow \sqrt {18} < \sqrt {20} < \sqrt {23} < \sqrt {25} \\
\Rightarrow 3\sqrt 2 < 2\sqrt 5 < \sqrt {23} < 5\\
b)\left\{ \begin{array}{l}
5\sqrt 2 = \sqrt {25.2} = \sqrt {50} \\
2\sqrt {13} = \sqrt {4.13} = \sqrt {52} \\
4\sqrt 3 = \sqrt {16.3} = \sqrt {48} \\
\sqrt {47} = \sqrt {47}
\end{array} \right.\\
\Rightarrow \sqrt {47} < \sqrt {48} < \sqrt {50} < \sqrt {52} \\
\Rightarrow \sqrt {47} < 4\sqrt 3 < 5\sqrt 2 < 2\sqrt {13}
\end{array}$
Bài 15:
$\begin{array}{l}
a)A = 4\sqrt {\dfrac{{25x}}{4}} - \dfrac{8}{3}\sqrt {\dfrac{{9x}}{4}} - \dfrac{4}{{3x}}\sqrt {\dfrac{{9{x^3}}}{{64}}} \left( {DK:x > 0} \right)\\
= 4\sqrt {\dfrac{{25}}{4}} .\sqrt x - \dfrac{8}{3}\sqrt {\dfrac{9}{4}} .\sqrt x - \dfrac{4}{{3x}}\sqrt {\dfrac{{9{x^2}}}{{64}}} .\sqrt x \\
= 4.\dfrac{5}{2}.\sqrt x - \dfrac{8}{3}.\dfrac{3}{2}.\sqrt x - \dfrac{4}{{3x}}.\dfrac{{3x}}{8}.\sqrt x \\
= \dfrac{{11}}{2}\sqrt x \\
b)B = \dfrac{y}{2} + \dfrac{3}{4}\sqrt {1 - 4y + 4{y^2}} - \dfrac{3}{2}\left( {DK:y \le \dfrac{1}{2}} \right)\\
= \dfrac{y}{2} + \dfrac{3}{4}\sqrt {{{\left( {2y - 1} \right)}^2}} - \dfrac{3}{2}\\
= \dfrac{y}{2} + \dfrac{3}{4}\left| {2y - 1} \right| - \dfrac{3}{2}\\
= \dfrac{y}{2} + \dfrac{3}{4}\left( {1 - 2y} \right) - \dfrac{3}{2}\\
= \dfrac{y}{2} + \dfrac{3}{4} - \dfrac{3}{2}y - \dfrac{3}{2}\\
= - y - \dfrac{3}{4}
\end{array}$
