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Tính :
a) \( – \dfrac{1}{2}\sqrt {108} – \dfrac{1}{{15}}\sqrt {75} – \dfrac{1}{{22}}\sqrt {363} + \sqrt {12} \);
b) \(2\sqrt {\dfrac{{27}}{2}} – \sqrt {\dfrac{{48}}{9}} – \dfrac{2}{5}\sqrt {\dfrac{{75}}{{18}}} \);
c) \(2y\sqrt {45} + 3\sqrt {20{y^2}} \);
d) \(3x\sqrt {72x} – 9\sqrt {50{x^3}} \) với \(x \ge 0\).
+) Sử dụng công thức: \(\sqrt {{A^2}B} = \left| A \right|\sqrt B = \left\{ \begin{array}{l}A\;\;khi\;\;A \ge 0\\ – A\;\;\;khi\;\;A < 0\end{array} \right..\)
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\(\begin{array}{l}a)\; – \dfrac{1}{2}\sqrt {108} – \dfrac{1}{{15}}\sqrt {75} – \dfrac{1}{{22}}\sqrt {363} + \sqrt {12} \\ = – \dfrac{1}{2}\sqrt {{6^2}.3} – \dfrac{1}{{15}}\sqrt {{5^2}.3} – \dfrac{1}{{22}}\sqrt {{{11}^2}.3} + \sqrt {{2^2}.3} \\ = – \dfrac{1}{2}.6\sqrt 3 – \dfrac{1}{{15}}.5\sqrt 3 – \dfrac{1}{{22}}.11\sqrt 3 + 2\sqrt 3 \\ = – \dfrac{{11\sqrt 3 }}{6}.\end{array}\) \(\begin{array}{l}c)\;2y\sqrt {45} + 3\sqrt {20{y^2}} \\ = 2y\sqrt {{3^2}.5} + 3\sqrt {{2^2}.5} .\sqrt {{y^2}} \\ = 6y\sqrt 5 + 6\sqrt 5 \left| y \right|\\ = \left\{ \begin{array}{l}6\sqrt 5 y + 6\sqrt 5 y\;\;\;\;khi\;\;\;y \ge 0\\6\sqrt 5 y – 6\sqrt 5 y\;\;\;\;khi\;\;y < 0\end{array} \right.\\ = \left\{ \begin{array}{l}12\sqrt 5 y\;\;\;khi\;\;y \ge 0\\0\;\;\;\;\;\;\;\;\;\;khi\;\;y < 0\end{array} \right..\end{array}\) |
\(\begin{array}{l}b)\;2\sqrt {\dfrac{{27}}{2}} – \sqrt {\dfrac{{48}}{9}} – \dfrac{2}{5}\sqrt {\dfrac{{75}}{{18}}} \\ = 2\sqrt {\dfrac{{{3^2}.3.2}}{{{2^2}}}} – \sqrt {\dfrac{{{4^2}.3}}{{{3^2}}}} – \dfrac{2}{5}\sqrt {\dfrac{{{5^2}.3}}{{{3^2}.2}}} \\ = 2.\dfrac{{3\sqrt 3 }}{2} – \dfrac{4}{3}\sqrt 3 – \dfrac{2}{5}.\dfrac{5}{3}\sqrt {\dfrac{3}{2}} \\ = 3\sqrt 3 – \dfrac{{4\sqrt 3 }}{3} – \dfrac{2}{3}\dfrac{{\sqrt 6 }}{2}\\ = \dfrac{{5\sqrt 3 }}{3} – \dfrac{{\sqrt 6 }}{3} = \dfrac{{4\sqrt 3 }}{3}.\end{array}\) \(\begin{array}{l}d)\;3x\sqrt {72x} – 9\sqrt {50{x^3}} \;\;\;\left( {x \ge 0} \right)\\ = 3x\sqrt {{6^2}.2x} – 9\sqrt {{5^2}.2.{x^2}.x} \\ = 3x.6\sqrt {2x} – 9.5.\left| x \right|\sqrt {2x} \\ = 18x\sqrt {2x} – 45x\sqrt {2x} \\ = – 27x\sqrt {2x} .\end{array}\) |