Bài 1. a. \(\sqrt {180{x^2}} = \sqrt {9.4.5{x^2}} = 6\left| x \right|\sqrt 5 \,\)\( = \left\{ {\matrix{ {6x\sqrt 5 \,\text{ nếu }\,x \ge 0} \cr { - 6x\sqrt 5 \,\text{ nếu }\,x < 0} \cr } } \right.\)
b. \(\sqrt {3{x^2} - 6xy + 3{y^2}} \)
\(= \sqrt {3{{\left( {x - y} \right)}^2}} \)
\(= \left| {x - y} \right|\sqrt 3 \)
\(= \left\{ {\matrix{ {\left( {x - y} \right)\sqrt 3 \,\text{ nếu }\,x \ge y} \cr {\left( {y - x} \right)\sqrt 3 \,\text{ nếu }\,x < y} \cr } } \right.\)
Bài 2. a. \({1 \over {xy}}.\sqrt {{{{x^2}{y^2}} \over 2}} = {{\left| {xy} \right|} \over {xy}}.\sqrt {{1 \over 2}} \)\(\,= \left\{ {\matrix{ {{1 \over {\sqrt 2 }}\,\text{ nếu}\,xy > 0} \cr { - {1 \over {\sqrt 2 }}\,\text{ nếu }\,xy < 0} \cr } } \right.\)
b. \({3 \over {{a^2} - {b^2}}}.\sqrt {{{2{{\left( {a + b} \right)}^2}} \over 9}} = {3 \over {3\left( {{a^2} - {b^2}} \right)}}\left| {a + b} \right|.\sqrt 2\)
\( = \left\{ {\matrix{ {{{\sqrt 2 } \over {a - b}}\,\text{ nếu }\,a + b > 0,a \ne b} \cr {{{\sqrt 2 } \over {b - a}}\,\text{ nếu }\,a + b < 0,a \ne b} \cr } } \right.\)
Bài 3. Ta có:
\(\left( * \right) \Leftrightarrow \sqrt {4\left( {x - 5} \right)} + \sqrt {x + 5} \)\(\, - {1 \over 3}\sqrt {9\left( {x - 5} \right)} = 4 \)
\(\eqalign{ & \Leftrightarrow 2\sqrt {x - 5} + \sqrt {x - 5} - \sqrt {x - 5} = 4 \cr & \Leftrightarrow \sqrt {x - 5} = 2 \Leftrightarrow x - 5 = 4 \cr&\Leftrightarrow x = 9 \cr} \)
