Đáp án:
$\begin{array}{l}
1)\left( {1 - \sqrt 2 } \right){x^2} - 2\left( {1 + \sqrt 2 } \right)x + 1 + 3\sqrt 2 = 0\\
\Rightarrow \left( {1 - \sqrt 2 } \right){x^2} - \left( {1 - \sqrt 2 } \right)x - \left( {1 + 3\sqrt 2 } \right)x + 1 + 3\sqrt 2 = 0\\
\Rightarrow \left( {1 - \sqrt 2 } \right)x\left( {x - 1} \right) - \left( {1 + 3\sqrt 2 } \right)\left( {x - 1} \right) = 0\\
\Rightarrow \left( {x - 1} \right)\left( {\left( {1 - \sqrt 2 } \right)x - 1 - 3\sqrt 2 } \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 1\\
x = \frac{{1 + 3\sqrt 2 }}{{1 - \sqrt 2 }} = - 7 - 4\sqrt 2
\end{array} \right.\\
2)3{x^2} - 19x - 22 = 0\\
\Leftrightarrow 3{x^2} + 3x - 22x - 22 = 0\\
\Leftrightarrow \left( {x + 1} \right).3x - 22\left( {x + 1} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {3x - 22} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 1 = 0\\
3x - 22 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 1\\
x = \frac{{22}}{3}
\end{array} \right.\\
3)5{x^2} + 24x + 19 = 0\\
\Leftrightarrow 5{x^2} + 5x + 19x + 19 = 0\\
\Leftrightarrow \left( {x + 1} \right).5x + 19\left( {x + 1} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {5x + 19} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 1\\
x = - \frac{{19}}{5}
\end{array} \right.
\end{array}$
