Đáp án:
$\begin{array}{l}
1)A = \left( {\sqrt 2 + \sqrt {3 - \sqrt 5 } } \right).\sqrt 2 \\
= \sqrt 2 .\sqrt 2 + \sqrt {3 - \sqrt 5 } .\sqrt 2 \\
= 2 + \sqrt {6 - 2\sqrt 5 } \\
= 2 + \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} \\
= 2 + \sqrt 5 - 1\\
= 1 + \sqrt 5 \\
2)A = \left( {3 + \sqrt 5 } \right).\left( {\sqrt {10} - \sqrt 2 } \right).\sqrt {3 - \sqrt 5 } \\
= \left( {3 + \sqrt 5 } \right).\left( {\sqrt 5 - 1} \right).\sqrt 2 .\sqrt {3 - \sqrt 5 } \\
= \dfrac{1}{2}.\left( {6 + 2\sqrt 5 } \right).\left( {\sqrt 5 - 1} \right).\sqrt {6 - 2\sqrt 5 } \\
= \dfrac{1}{2}.{\left( {\sqrt 5 + 1} \right)^2}.\left( {\sqrt 5 - 1} \right).\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} \\
= \dfrac{1}{2}.{\left( {\sqrt 5 + 1} \right)^2}.{\left( {\sqrt 5 - 1} \right)^2}\\
= \dfrac{1}{2}.{\left( {5 - 1} \right)^2}\\
= 8\\
B2)\\
1)\sqrt {49{x^3}} \left( {x \le 0} \right)\\
Dkxd:49{x^3} \ge 0\\
\Rightarrow x \ge 0\\
\Rightarrow x = 0\\
\Rightarrow \sqrt {49{x^3}} = 0\\
2)\sqrt {{x^4}{{\left( {4 - x} \right)}^2}} \left( {x \ge 4} \right)\\
= {x^2}.\left| {4 - x} \right|\\
= {x^2}.\left( {x - 4} \right)\\
= {x^3} - 4{x^2}\\
3)\sqrt {20.45.{{\left( {x - 7} \right)}^3}} \left( {x \ge 7} \right)\\
= 2\sqrt 5 .3\sqrt 5 .\left( {x - 7} \right)\sqrt {x - 7} \\
= 30\left( {x - 7} \right)\sqrt {x - 7}
\end{array}$
