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Đưa thừa số ra ngoài dấu căn :
a) \(\sqrt {32} ;\;2\sqrt {75} ;\;3\sqrt {80} ;\;\dfrac{5}{6}\sqrt {48} ;\;\sqrt {108} \).
b) \(\sqrt {63{a^2}} \left( {a \ge 0} \right)\); \(2\sqrt {12a{b^2}} \left( {a \ge 0,b < 0} \right)\); \(\sqrt {125{a^2}{b^2}} \left( {ab \ge 0} \right)\).
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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p style=”text-align: justify;”>\(\begin{array}{l}a)\;\;\sqrt {32} = \sqrt {{4^2}.2} = 4\sqrt 2 \\ 2\sqrt {75} = 2\sqrt {{5^2}.3} = 2.5\sqrt 3 = 10\sqrt 3 \\3\sqrt {80} = 3\sqrt {{4^2}.5} = 3.4\sqrt 5 = 12\sqrt 5 \\ \dfrac{5}{6}\sqrt {48} = \dfrac{5}{6}.\sqrt {{4^2}.3} = \dfrac{5}{6}.4\sqrt 3 = \dfrac{{10\sqrt 3 }}{3}\\\sqrt {108} = \sqrt {{6^2}.3} = 6\sqrt 3 .\end{array}\)
\(\eqalign{
& b)\,\sqrt {63{a^2}} = \sqrt {9.7.{a^2}} = 3a\sqrt 7 \left( {a \geqslant 0} \right) \cr
& 2\sqrt {12a{b^2}} = 2\sqrt {4.3.a{b^2}} = – 4b\sqrt {3a} \left( {a \geqslant 0,b < 0} \right) \cr
& \sqrt {125{a^2}{b^2}} = \sqrt {{5^3}{a^2}{b^2}} = 5ab\sqrt 5 \left( {ab \geqslant 0} \right) \cr} \)