Đáp án:

\(\begin{array}{l}
11,\,\,\,\,\left[ \begin{array}{l}
x = \sqrt 2  - 1 - \sqrt 3 \\
x = 1 - \sqrt 2  - \sqrt 3 
\end{array} \right.\\
13,\,\,\,\,\left[ \begin{array}{l}
x =  - \dfrac{1}{2}\\
x = \dfrac{1}{2} - \sqrt 7 
\end{array} \right.\\
14,\,\,\,\,\left[ \begin{array}{l}
x = 1\\
x = \dfrac{{2\sqrt {10}  - 5}}{5}
\end{array} \right.\\
15,\,\,\,\,\left[ \begin{array}{l}
x = 1\\
x =  - \left( {3\sqrt 2  + 1} \right)
\end{array} \right.\\
16,\,\,\,\,\left[ \begin{array}{l}
x =  - 1\\
x = \dfrac{{5 - 2\sqrt {30} }}{5}
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
11,\\
\sqrt {3 - 2\sqrt 2 }  - \sqrt {{x^2} + 2x\sqrt 3  + 3}  = 0\\
 \Leftrightarrow \sqrt {2 - 2\sqrt 2  + 1}  - \sqrt {{x^2} + 2.x.\sqrt 3  + {{\sqrt 3 }^2}}  = 0\\
 \Leftrightarrow \sqrt {{{\sqrt 2 }^2} - 2.\sqrt 2 .1 + {1^2}}  - \sqrt {{x^2} + 2.x.\sqrt 3  + {{\sqrt 3 }^2}}  = 0\\
 \Leftrightarrow \sqrt {{{\left( {\sqrt 2  - 1} \right)}^2}}  - \sqrt {{{\left( {x + \sqrt 3 } \right)}^2}}  = 0\\
 \Leftrightarrow \left| {\sqrt 2  - 1} \right| - \left| {x + \sqrt 3 } \right| = 0\\
 \Leftrightarrow \left| {x + \sqrt 3 } \right| = \left| {\sqrt 2  - 1} \right|\\
 \Leftrightarrow \left| {x + \sqrt 3 } \right| = \sqrt 2  - 1\\
 \Leftrightarrow \left[ \begin{array}{l}
x + \sqrt 3  = \sqrt 2  - 1\\
x + \sqrt 3  = 1 - \sqrt 2 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \sqrt 2  - 1 - \sqrt 3 \\
x = 1 - \sqrt 2  - \sqrt 3 
\end{array} \right.\\
13,\\
\sqrt {4{x^2} + 4x\sqrt 7  + 7}  - \sqrt {8 - 2\sqrt 7 }  = 0\\
 \Leftrightarrow \sqrt {{{\left( {2x} \right)}^2} + 2.2x.\sqrt 7  + {{\sqrt 7 }^2}}  - \sqrt {7 - 2\sqrt 7  + 1}  = 0\\
 \Leftrightarrow \sqrt {{{\left( {2x} \right)}^2} + 2.2x.\sqrt 7  + {{\sqrt 7 }^2}}  - \sqrt {{{\sqrt 7 }^2} - 2.\sqrt 7 .1 + {1^2}}  = 0\\
 \Leftrightarrow \sqrt {{{\left( {2x + \sqrt 7 } \right)}^2}}  - \sqrt {{{\left( {\sqrt 7  - 1} \right)}^2}}  = 0\\
 \Leftrightarrow \left| {2x + \sqrt 7 } \right| - \left| {\sqrt 7  - 1} \right| = 0\\
 \Leftrightarrow \left| {2x + \sqrt 7 } \right| = \left| {\sqrt 7  - 1} \right|\\
 \Leftrightarrow \left| {2x + \sqrt 7 } \right| = \sqrt 7  - 1\\
 \Leftrightarrow \left[ \begin{array}{l}
2x + \sqrt 7  = \sqrt 7  - 1\\
2x + \sqrt 7  = 1 - \sqrt 7 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x =  - 1\\
2x = 1 - 2\sqrt 7 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{1}{2}\\
x = \dfrac{1}{2} - \sqrt 7 
\end{array} \right.\\
14,\\
\sqrt {7 - 2\sqrt {10} }  - \sqrt {5{x^2} - 2x\sqrt {10}  + 2}  = 0\\
 \Leftrightarrow \sqrt {5 - 2\sqrt {10}  + 2}  - \sqrt {5{x^2} - 2x.\sqrt 5 .\sqrt 2  + 2}  = 0\\
 \Leftrightarrow \sqrt {{{\sqrt 5 }^2} - 2.\sqrt 5 .\sqrt 2  + {{\sqrt 2 }^2}}  - \sqrt {{{\left( {\sqrt 5 x} \right)}^2} - 2.\sqrt 5 x.\sqrt 2  + {{\sqrt 2 }^2}}  = 0\\
 \Leftrightarrow \sqrt {{{\left( {\sqrt 5  - \sqrt 2 } \right)}^2}}  - \sqrt {{{\left( {\sqrt 5 x - \sqrt 2 } \right)}^2}}  = 0\\
 \Leftrightarrow \left| {\sqrt 5  - \sqrt 2 } \right| - \left| {\sqrt 5 x - \sqrt 2 } \right| = 0\\
 \Leftrightarrow \left| {\sqrt 5 x - \sqrt 2 } \right| = \left| {\sqrt 5  - \sqrt 2 } \right|\\
 \Leftrightarrow \left| {\sqrt 5 x - \sqrt 2 } \right| = \sqrt 5  - \sqrt 2 \\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt 5 .x - \sqrt 2  = \sqrt 5  - \sqrt 2 \\
\sqrt 5 .x - \sqrt 2  = \sqrt 2  - \sqrt 5 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt 5 .x = \sqrt 5 \\
\sqrt 5 .x = 2\sqrt 2  - \sqrt 5 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \dfrac{{2\sqrt 2 }}{{\sqrt 5 }} - 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \dfrac{{2\sqrt {10} }}{5} - 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \dfrac{{2\sqrt {10}  - 5}}{5}
\end{array} \right.\\
15,\\
\sqrt {11 + 6\sqrt 2 }  = \sqrt {2{x^2} - 6x\sqrt 2  + 9} \\
 \Leftrightarrow \sqrt {9 + 6\sqrt 2  + 2}  = \sqrt {{{\left( {\sqrt 2 x} \right)}^2} + 2.\sqrt 2 x.3 + {3^2}} \\
 \Leftrightarrow \sqrt {{3^2} + 2.3.\sqrt 2  + {{\sqrt 2 }^2}}  = \sqrt {{{\left( {\sqrt 2 .x + 3} \right)}^2}} \\
 \Leftrightarrow \sqrt {{{\left( {3 + \sqrt 2 } \right)}^2}}  = \left| {\sqrt 2 x + 3} \right|\\
 \Leftrightarrow 3 + \sqrt 2  = \left| {\sqrt 2 .x + 3} \right|\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt 2 .x + 3 = 3 + \sqrt 2 \\
\sqrt 2 .x + 3 =  - 3 - \sqrt 2 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt 2 x = \sqrt 2 \\
\sqrt 2 x =  - 6 - \sqrt 2 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x =  - \dfrac{{6 + \sqrt 2 }}{{\sqrt 2 }}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x =  - \dfrac{{6\sqrt 2  + 2}}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x =  - \left( {3\sqrt 2  + 1} \right)
\end{array} \right.\\
16,\\
\sqrt {120}  = \sqrt {4.30}  = 2\sqrt {30} \\
\sqrt {11 - \sqrt {120} }  = \sqrt {5{x^2} + x\sqrt {120}  + 6} \\
 \Leftrightarrow \sqrt {6 - 2\sqrt {30 + 5} }  = \sqrt {5{x^2} + 2x.\sqrt {30}  + 6} \\
 \Leftrightarrow \sqrt {{{\sqrt 6 }^2} - 2.\sqrt 6 .\sqrt 5  + {{\sqrt 5 }^2}}  = \sqrt {{{\left( {\sqrt 5 x} \right)}^2} + 2.\sqrt 5 x.\sqrt 6  + {{\sqrt 6 }^2}} \\
 \Leftrightarrow \sqrt {{{\left( {\sqrt 6  - \sqrt 5 } \right)}^2}}  = \sqrt {{{\left( {\sqrt 5 x + \sqrt 6 } \right)}^2}} \\
 \Leftrightarrow \left| {\sqrt 6  - \sqrt 5 } \right| = \left| {\sqrt 5 x + \sqrt 6 } \right|\\
 \Leftrightarrow \left| {\sqrt 5 x + \sqrt 6 } \right| = \sqrt 6  - \sqrt 5 \\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt 5 x + \sqrt 6  = \sqrt 6  - \sqrt 5 \\
\sqrt 5 x + \sqrt 6  = \sqrt 5  - \sqrt 6 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt 5 .x =  - \sqrt 5 \\
\sqrt 5 .x = \sqrt 5  - 2\sqrt 6 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 1\\
x = \dfrac{{\sqrt 5  - 2\sqrt 6 }}{{\sqrt 5 }}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 1\\
x = \dfrac{{5 - 2\sqrt {30} }}{5}
\end{array} \right.
\end{array}\)