Đáp án:
$\begin{array}{l}
a)\sqrt {x - 3} = 6\left( {dkxd:x \ge 3} \right)\\
\Rightarrow x - 3 = {6^2}\\
\Rightarrow x = 36 + 3\\
\Rightarrow x = 39\left( {tmdk} \right)\\
Vậy\,x = 39\\
b)\sqrt {3 - x} = 6\left( {dkxd:x \le 3} \right)\\
\Rightarrow 3 - x = 36\\
\Rightarrow x = - 33\left( {tmdk} \right)\\
Vậy\,x = - 33\\
c)\sqrt {x - 1} = 1 - x\\
Đkxđ:\left\{ \begin{array}{l}
x - 1 \ge 0\\
1 - x \ge 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ge 1\\
x \le 1
\end{array} \right. \Rightarrow x = 1\\
Vậy\,x = 1\\
d){x^4} - 2{x^2} - 8 = 0\\
\Rightarrow {x^4} - 4{x^2} + 2{x^2} - 8 = 0\\
\Rightarrow \left( {{x^2} - 4} \right).{x^2} + 2\left( {{x^2} - 4} \right) = 0\\
\Rightarrow \left( {{x^2} - 4} \right)\left( {{x^2} + 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
{x^2} = 4\\
{x^2} = - 2\left( {ktm} \right)
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 2\\
x = - 2
\end{array} \right.\\
Vậy\,x = 2\,hoặc\,x = - 2\\
g)\\
C1:\left\{ \begin{array}{l}
x + y = 3\\
3x - y = 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left( {x + y} \right) + \left( {3x - y} \right) = 3 + 1\\
x + y = 3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
4x = 4\\
y = 3 - x
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 1\\
y = 3 - 1 = 2
\end{array} \right.\\
Vậy\left( {x;y} \right) = \left( {1;2} \right)\\
C2:\left\{ \begin{array}{l}
x + y = 3\\
3x - y = 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x + y = 3\\
y = 3x - 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x + 3x - 1 = 3\\
y = 3x - 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
4x = 4\\
y = 3x - 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 1\\
y = 2
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left( {1;2} \right)
\end{array}$
