Đáp án:
$\begin{array}{l}
b)\text{Đặt}:{x^2} = a\left( {a \ge 0} \right)\\
\Rightarrow - {a^2} + 10a - 9 = 0\\
\Rightarrow {a^2} - 10a + 9 = 0\\
\Rightarrow \left( {a - 1} \right)\left( {a - 9} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
a = 1\\
a = 9
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
{x^2} = 1\\
{x^2} = 9
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \pm 1\\
x = \pm 3
\end{array} \right.\\
\text{Vậy}\,x \in \left\{ { - 3; - 1;1;3} \right\}\\
c)\text{Đặt}:{x^2} = a\left( {a \ge 0} \right)\\
\Rightarrow {a^2} - 3a - 4 = 0\\
\Rightarrow {a^2} - 4a + a - 4 = 0\\
\Rightarrow \left( {a - 4} \right)\left( {a + 1} \right) = 0\\
\Rightarrow a = 4\left( {do:a \ge 0} \right)\\
\Rightarrow {x^2} = 4\\
\Rightarrow \left[ \begin{array}{l}
x = 2\\
x = - 2
\end{array} \right.\\
\text{Vậy}\,x = 2;x = - 2\\
e){a^2} - a + 3 = 0\\
\Rightarrow {a^2} - 2.a.\frac{1}{2} + \frac{1}{4} + \frac{{11}}{4} = 0\\
\Rightarrow {\left( {a - \frac{1}{2}} \right)^2} + \frac{{11}}{4} = 0\left( {vn} \right)\\
\text{Vậy pt vô nghiệm}\\
f)\left( {1 - a} \right)\left( {1 + a} \right) + 3 = 0\\
\Rightarrow 1 - {a^2} + 3 = 0\\
\Rightarrow {a^2} = 4\\
\Rightarrow \left[ \begin{array}{l}
a = 2\left( {tm} \right)\\
a = - 2\left( {ktm} \right)
\end{array} \right.\\
\Rightarrow {x^2} = 2\\
\Rightarrow \left[ \begin{array}{l}
x = \sqrt 2 \\
x = - \sqrt 2
\end{array} \right.\\
\text{Vậy}\,x = \sqrt 2 ;x = - \sqrt 2
\end{array}$
