Đáp án:
$\begin{array}{l}
4{a^2}x + 4a - {b^2}x - b = 0\\
\Rightarrow \left( {4{a^2} - {b^2}} \right)x = b - 4a\\
+ )neu\left\{ \begin{array}{l}
4{a^2} - {b^2} = 0\\
b - 4a = 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left( {2a - b} \right)\left( {2a + b} \right) = 0\\
4a = b
\end{array} \right. \Rightarrow a = b = 0\\
thi\,pt \Leftrightarrow 0.x = 0\,\forall x\\
Vay\,\,a = b = 0\,thi\,pt\,dung\,\forall x\\
+ )\,neu\left\{ \begin{array}{l}
4{a^2} - {b^2} \ne 0\\
b - 4a = 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
b \ne 2a;b \ne - 2a\\
b = 4a
\end{array} \right. \Rightarrow a \ne 0\\
thi\,pt\,co\,nghiem\,duy\,nhat\,x = 0\\
+ )\,neu\left\{ \begin{array}{l}
4{a^2} - {b^2} = 0\\
b - 4a \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
b = 2a;hoac\,\,b = - 2a\\
b \ne 4a
\end{array} \right. \Rightarrow b \ne 0\\
thi\,pt\,vo\,nghiem\\
+ )\,neu\left\{ \begin{array}{l}
4{a^2} - {b^2} \ne 0\\
b - 4a \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
b \ne 2a;hoac\,\,b \ne - 2a\\
b \ne 4a
\end{array} \right. \Rightarrow a \ne 0;b \ne 0\\
thi\,pt\,co\,co\,nghiem\,:x = \frac{{b - 4a}}{{4{a^2} - {b^2}}}
\end{array}$
