Đáp án:
$\begin{array}{l}
6.1\\
a)\sqrt {96.125} = \sqrt {16.6.25.5} = 4.5\sqrt {30} = 20\sqrt {30} \\
b)\sqrt {{a^4}{b^5}} = {a^2}{b^2}\sqrt b \\
c)\sqrt {{a^6}{b^{11}}} = {a^3}{b^5}\sqrt b \\
d)\sqrt {{a^3}{{\left( {1 - a} \right)}^4}} = {\left( {1 - a} \right)^2}.a\sqrt a \\
6.2\\
a)x\sqrt {13} = \sqrt {{x^2}.13} = \sqrt {13{x^2}} \left( {x \ge 0} \right)\\
b)x\sqrt 2 = - \sqrt {{x^2}.2} = - \sqrt {2{x^2}} \left( {x < 0} \right)\\
c)x\sqrt {\dfrac{{ - 11}}{x}} = - \sqrt {{x^2}.\dfrac{{ - 11}}{x}} = - \sqrt { - 11x} \\
6.3\\
a)4\sqrt 7 = \sqrt {16.7} = \sqrt {112} \\
3\sqrt {13} = \sqrt {9.13} = \sqrt {117} > \sqrt {112} \\
\Leftrightarrow 4\sqrt 7 < 3\sqrt {13} \\
b)\dfrac{1}{4}\sqrt {82} = \sqrt {\dfrac{{82}}{{16}}} = \sqrt {\dfrac{{41}}{4}} \\
6\sqrt {\dfrac{1}{7}} = \sqrt {\dfrac{{36}}{7}} < \sqrt {\dfrac{{41}}{4}} \\
\Leftrightarrow \dfrac{1}{4}\sqrt {82} > 6\sqrt {\dfrac{1}{7}} \\
6.4)\\
a)\sqrt {98} - \sqrt {72} + 0,5\sqrt 8 \\
= 7\sqrt 2 - 6\sqrt 2 + 0,5.2\sqrt 2 \\
= \sqrt 2 + \sqrt 2 \\
= 2\sqrt 2 \\
b)\sqrt {16a} + 2\sqrt {40a} - 3\sqrt {90a} \\
= 4\sqrt a + 2.2\sqrt {10a} - 3.3\sqrt {10a} \\
= 4\sqrt a - 5\sqrt {10a} \\
c)\left( {2\sqrt 3 + \sqrt 5 } \right).\sqrt 3 - \sqrt {60} \\
= 2.3 + \sqrt {15} - 4\sqrt {15} \\
= 6 - 3\sqrt {15} \\
d)\left( {\sqrt {99} - \sqrt {18} - \sqrt {11} } \right).\sqrt {11} + 3\sqrt {22} \\
= \left( {3\sqrt {11} - 3\sqrt 2 - \sqrt {11} } \right).\sqrt {11} + 3\sqrt {22} \\
= 2.\sqrt {11} .\sqrt {11} - 3\sqrt {22} + 3\sqrt {22} \\
= 22\\
6,5\\
a)\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)\\
= {\left( {\sqrt x } \right)^3} - 1\\
= x\sqrt x - 1\\
b)\left( {\sqrt x + \sqrt y } \right)\left( {x - \sqrt {xy} + y} \right)\\
= {\left( {\sqrt x } \right)^3} + {\left( {\sqrt y } \right)^3}\\
= x\sqrt x + y\sqrt y \\
c)\left( {2\sqrt x + \sqrt y } \right)\left( {3\sqrt x - 2\sqrt y } \right)\\
= 6x - 4\sqrt {xy} + 3\sqrt {xy} - 2y\\
= 6x - \sqrt {xy} + 2y\\
6.6)\\
a)\dfrac{{\left( {x\sqrt y + y\sqrt x } \right)\left( {\sqrt x - \sqrt y } \right)}}{{\sqrt {xy} }}\\
= \dfrac{{\sqrt {xy} \left( {\sqrt x + \sqrt y } \right)\left( {\sqrt x - \sqrt y } \right)}}{{\sqrt {xy} }}\\
= x - y\\
b)x + 2\sqrt {3x - 9} \\
= x - 3 + 2.\sqrt 3 .\sqrt {x - 3} + 3\\
= {\left( {\sqrt {x - 3} + \sqrt 3 } \right)^2}
\end{array}$
