Rút gọn biểu thức :
a) \(\sqrt {0,16{x^2}} \) với \(x < 0\);
b) \(\sqrt {12.75{{\left( {2 - x} \right)}^2}} \) với \(x < 2\);
c) \(\sqrt {\dfrac{x}{{15}}} .\sqrt {\dfrac{{5x}}{3}} \) với \(x > 0\);
d) \(\sqrt {3x} .\sqrt {\dfrac{{12}}{x}} \) với \(x > 0\).
+) Sử dụng công thức; \(\sqrt {{A^2}} = \left| A \right| = \left\{ \begin{array}{l}A\;\;khi\;\;A \ge 0\\ - A\;\;khi\;A < 0\end{array} \right.\;\;\sqrt A .\sqrt B = \sqrt {AB} ,\;\;\sqrt {\dfrac{A}{B}} = \dfrac{{\sqrt A }}{{\sqrt B }}.\)
\(\begin{array}{l}a)\;\sqrt {0,16{x^2}} = \sqrt {0,16} .\sqrt {{x^2}} = 0,4.\left| x \right| = - 0,4x\;\;\;\left( {do\;\;x < 0} \right).\\b)\;\;\sqrt {12.75{{\left( {2 - x} \right)}^2}} = \sqrt {4.3.3.25{{\left( {2 - x} \right)}^2}} \\ = \sqrt 4 .\sqrt {{3^2}} .\sqrt {25} .\sqrt {{{\left( {2 - x} \right)}^2}} = 2.3.5.\left| {2 - x} \right|\\ = 30\left( {2 - x} \right)\;\;\;\;\left( {do\;\;x < 2 \Rightarrow \left| {2 - x} \right| = 2 - x} \right).\\c)\;\sqrt {\dfrac{x}{{15}}} .\sqrt {\dfrac{{5x}}{3}} = \sqrt {\dfrac{x}{{15}}.\dfrac{{5x}}{3}} = \sqrt {\dfrac{{{x^2}}}{9}} = \dfrac{{\sqrt {{x^2}} }}{{\sqrt 9 }} = \dfrac{x}{3}\;\;\;\left( {do\;\;x > 0} \right).\\d)\;\;\sqrt {3x} .\sqrt {\dfrac{{12}}{x}} = \sqrt {3x.\dfrac{{12}}{x}} = \sqrt {3.12} = \sqrt {36} = 6.\;\end{array}\)