Đáp án:
$\begin{array}{l}
a){x^2} - 2\sqrt 5 x + 5 = 0\\
\Rightarrow {\left( {x - \sqrt 5 } \right)^2} = 0\\
\Rightarrow x = \sqrt 5 \\
b){x^2} - 6x + 14 = 0\\
\Rightarrow {x^2} - 6x + 9 + 5 = 0\\
\Rightarrow {\left( {x - 3} \right)^2} + 5 = 0\left( {vô\,nghiệm} \right)\\
c)2\sqrt 3 {x^2} + x + 1 = \sqrt 3 x + 1\\
\Rightarrow 2\sqrt 3 {x^2} + x - \sqrt 3 x = 0\\
\Rightarrow x.\left( {2\sqrt 3 x + 1 - \sqrt 3 } \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 3 }} = \dfrac{{3 - \sqrt 3 }}{6}
\end{array} \right.\\
d)16{x^2} - 40x + 25 = 0\\
\Rightarrow {\left( {4x} \right)^2} - 2.4x.5 + {5^2} = 0\\
\Rightarrow {\left( {4x - 5} \right)^2} = 0\\
\Rightarrow x = \dfrac{5}{4}\\
e){x^2} + 2\sqrt 2 x + 4 = 3x + \sqrt 2 \\
\Rightarrow {x^2} + \left( {2\sqrt 2 - 3} \right).x + 4 - \sqrt 2 = 0\\
\Rightarrow {x^2} + 2.\left( {\sqrt 2 - \dfrac{3}{2}} \right).x + {\left( {\sqrt 2 - \dfrac{3}{2}} \right)^2}\\
- {\left( {\sqrt 2 - \dfrac{3}{2}} \right)^2} + 4 - \sqrt 2 = 0\\
\Rightarrow {\left( {x + \sqrt 2 - \dfrac{3}{2}} \right)^2} + \dfrac{{33 - 16\sqrt 2 }}{4} = 0\\
\Rightarrow pt\,vô\,nghiệm
\end{array}$
