Đáp án:
$\begin{array}{l}
11)a)\sqrt {15 + 6\sqrt 6 } \\
= \sqrt {9 + 2.3.\sqrt 6 + 6} \\
= \sqrt {{{\left( {3 + \sqrt 6 } \right)}^2}} \\
= 3 + \sqrt 6 \\
b)\sqrt {29 - 4\sqrt 7 } \\
= \sqrt {28 - 2.2.\sqrt 7 + 1} \\
= \sqrt {{{\left( {2\sqrt 7 - 1} \right)}^2}} \\
= 2\sqrt 7 - 1\\
c)\sqrt {2\left( {4 - \sqrt 7 } \right)} = \sqrt {8 - 2\sqrt 7 } \\
= \sqrt {{{\left( {\sqrt 7 - 1} \right)}^2}} = \sqrt 7 - 1\\
d)\sqrt {17 - 12\sqrt 2 } + \sqrt {17 + 12\sqrt 2 } \\
= \sqrt {9 - 2.3.2\sqrt 2 + 8} + \sqrt {9 + 2.3.2\sqrt 2 + 8} \\
= \sqrt {{{\left( {3 - 2\sqrt 2 } \right)}^2}} + \sqrt {{{\left( {3 + 2\sqrt 2 } \right)}^2}} \\
= 3 - 2\sqrt 2 + 3 + 2\sqrt 2 \\
= 6\\
e)\sqrt {x - 1 - 2\sqrt {x - 2} } \\
= \sqrt {x - 2 - 2\sqrt {x - 2} + 1} \\
= \sqrt {{{\left( {\sqrt {x - 2} - 1} \right)}^2}} \\
= \left| {\sqrt {x - 2} - 1} \right|\\
f)\sqrt {x + 2 - 4\sqrt {x - 2} } \\
= \sqrt {x - 2 - 4\sqrt {x - 2} + 4} \\
= \sqrt {{{\left( {\sqrt {x - 2} - 2} \right)}^2}} \\
= \left| {\sqrt {x - 2} - 2} \right|\\
12)\\
a)\sqrt {1 - 4x + 4{x^2}} = x - 3\left( {dkxk:x \ge 3} \right)\\
\Rightarrow \sqrt {{{\left( {2x - 1} \right)}^2}} = x - 3\\
\Rightarrow 2x - 1 = x - 3\left( {do:x \ge 3 \Rightarrow 2x - 1 > 0} \right)\\
\Rightarrow x = - 2\left( {ktm} \right)\\
b)\sqrt {x - 2 + 2\sqrt {x - 3} } + \sqrt {x + 6 + 6\sqrt {x - 3} } = 4\\
\left( {dkxd:x \ge 3} \right)\\
\Rightarrow \sqrt {x - 3 + 2\sqrt {x - 3} + 1} \\
+ \sqrt {x - 3 + 6\sqrt {x - 3} + 9} = 4\\
\Rightarrow \sqrt {{{\left( {\sqrt {x - 3} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {x - 3} + 3} \right)}^2}} = 4\\
\Rightarrow \sqrt {x - 3} + 1 + \sqrt {x - 3} + 3 = 4\\
\Rightarrow \sqrt {x - 3} = 0\\
\Rightarrow x = 3\left( {tmdk} \right)\\
13)P = \sqrt {9{x^2} - 6x + 1} + 1 - 4x\\
= \sqrt {{{\left( {3x - 1} \right)}^2}} + 1 - 4x\\
= 3x - 1 + 1 - 4x\left( {do:x > 1} \right)\\
= - x\\
P = - 4\\
\Rightarrow - x = - 4\\
\Rightarrow x = 4\left( {tmdk} \right)
\end{array}$
