Đáp án:
$\begin{array}{l}
1)\sqrt {x - 2 - 2\sqrt {x - 3} } .\sqrt {x + 1 - 4\sqrt {x - 3} } \\
= \sqrt {x - 3 - 2.\sqrt {x - 3} + 1} .\\
\sqrt {x - 3 - 4\sqrt {x - 3} + 4} \\
= \sqrt {{{\left( {\sqrt {x - 3} - 1} \right)}^2}} .\sqrt {{{\left( {\sqrt {x - 3} - 2} \right)}^2}} \\
= \left| {\sqrt {x - 3} - 1} \right|.\left| {\sqrt {x - 3} - 2} \right|\\
= \left| {x - 3 - 3\sqrt {x - 3} + 2} \right|\\
= \left| {x - 1 - 3\sqrt {x - 3} } \right|\\
2)\left( {4\sqrt 3 - 4} \right).\sqrt {3 + \sqrt {5 - \sqrt {13 + 2\sqrt {12} } } } \\
= 4.\left( {\sqrt 3 - 1} \right).\sqrt {3 + \sqrt {5 - \sqrt {{{\left( {\sqrt {12} + 1} \right)}^2}} } } \\
= 4\left( {\sqrt 3 - 1} \right).\sqrt {3 + \sqrt {5 - \sqrt {12} - 1} } \\
= 4\left( {\sqrt 3 - 1} \right).\sqrt {3 + \sqrt {4 - 2\sqrt 3 } } \\
= 4\left( {\sqrt 3 - 1} \right).\sqrt {3 + \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} } \\
= 4\left( {\sqrt 3 - 1} \right).\sqrt {3 + \sqrt 3 - 1} \\
= 4\left( {\sqrt 3 - 1} \right).\sqrt {2 + \sqrt 3 } \\
= 2\sqrt 2 .\left( {\sqrt 3 - 1} \right).\sqrt 2 .\sqrt {2 + \sqrt 3 } \\
= 2\sqrt 2 \left( {\sqrt 3 - 1} \right).\sqrt {4 + 2\sqrt 3 } \\
= 2\sqrt 2 .\left( {\sqrt 3 - 1} \right).\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} \\
= 2\sqrt 2 \left( {\sqrt 3 - 1} \right).\left( {\sqrt 3 + 1} \right)\\
= 2\sqrt 2 \left( {3 - 1} \right)\\
= 2\sqrt 2 .2\\
= 4\sqrt 2 \\
3)\sqrt {0,25\sqrt {961} + 2\sqrt {10} + \sqrt {15} + \sqrt 6 }
\end{array}$
