Đáp án:

$\begin{array}{l}
1)a)\\
2{x^2} + 5x + 3 = 0\\
 \Rightarrow 2{x^2} + 2x + 3x + 3 = 0\\
 \Rightarrow \left( {x + 1} \right)\left( {2x + 3} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x =  - 1\\
x =  - \frac{3}{2}
\end{array} \right.\\
b){x^4} + 8{x^2} - 9 = 0\\
 \Rightarrow {x^4} - {x^2} + 9{x^2} - 9 = 0\\
 \Rightarrow \left( {{x^2} - 1} \right)\left( {{x^2} + 9} \right) = 0\\
 \Rightarrow {x^2} - 1 = 0\left( {do:{x^2} + 9 > 0\forall x} \right)\\
 \Rightarrow {x^2} = 1\\
 \Rightarrow \left[ \begin{array}{l}
x = 1\\
x =  - 1
\end{array} \right.\\
c)\left\{ \begin{array}{l}
x + 2y = 5\\
3x - y = 1
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
3x + 6y = 15\\
3x - y = 1
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
7y = 14\\
x + 2y = 5
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
y = 2\\
x = 5 - 2y = 1
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left( {1;2} \right)\\
d)\left( {{x^2} - 3} \right)\left( {3{x^2} + 6x} \right) = 0\\
 \Rightarrow \left( {{x^2} - 3} \right).3x.\left( {x + 2} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
{x^2} = 3\\
x = 0\\
x + 2 = 0
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \sqrt 3 \\
x =  - \sqrt 3 \\
x = 0\\
x =  - 2
\end{array} \right.\\
2)a){x^2} + 4x + 3 = 0\\
 \Rightarrow \left( {x + 1} \right)\left( {x + 3} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x =  - 1\\
x =  - 3
\end{array} \right.\\
b){x^4} - 4{x^2} = 0\\
 \Rightarrow {x^2}\left( {{x^2} - 4} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} = 4
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = 0\\
x = 2\\
x =  - 2
\end{array} \right.\\
c)\left\{ \begin{array}{l}
x + 6y = 3\\
x - 3y = 21
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
9y =  - 18\\
x = 3 - 6y
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
y =  - 2\\
x = 15
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left( {15; - 2} \right)\\
d)3{x^2} - 2\sqrt 6 x + 2 = 0\\
 \Rightarrow {\left( {\sqrt 3 x} \right)^2} - 2.\sqrt 3 x.\sqrt 2  + {\left( {\sqrt 2 } \right)^2} = 0\\
 \Rightarrow {\left( {\sqrt 3 x - \sqrt 2 } \right)^2} = 0\\
 \Rightarrow \sqrt 3 x = \sqrt 2 \\
 \Rightarrow x = \frac{{\sqrt 2 }}{{\sqrt 3 }} = \frac{{\sqrt 6 }}{3}
\end{array}$