Đáp án:
$\begin{array}{l}
a){\left( {x + 1} \right)^4} = {\left( {x + 1} \right)^2}\\
\Leftrightarrow {\left( {x + 1} \right)^4} - {\left( {x + 1} \right)^2} = 0\\
\Leftrightarrow {\left( {x + 1} \right)^2}.\left[ {{{\left( {x + 1} \right)}^2} - 1} \right] = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{\left( {x + 1} \right)^2} = 0\\
{\left( {x + 1} \right)^2} = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x + 1 = 0\\
x + 1 = 1\\
x + 1 = - 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 1\\
x = 0\\
x = - 2
\end{array} \right.\\
Vậy\,x = - 2;x = - 1;x = 0\\
b){\left( {x + 2} \right)^6} = {\left( {x + 2} \right)^4}\\
\Leftrightarrow {\left( {x + 2} \right)^4}.\left[ {{{\left( {x + 2} \right)}^2} - 1} \right] = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{\left( {x + 2} \right)^4} = 0\\
{\left( {x + 2} \right)^2} = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x + 2 = 0\\
x + 2 = 1\\
x + 2 = - 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 2\\
x = - 1\\
x = - 3
\end{array} \right.\\
Vậy\,x = - 3;x = - 2;x = - 1\\
c){\left( {x - 1} \right)^3} = {\left( {x - 1} \right)^2}\\
\Leftrightarrow {\left( {x - 1} \right)^2}.\left( {x - 1 - 1} \right) = 0\\
\Leftrightarrow {\left( {x - 1} \right)^2}.\left( {x - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
x - 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 2
\end{array} \right.\\
Vậy\,x = 1;x = 2\\
d){\left( {x - 3} \right)^6} = {\left( {x - 3} \right)^3}\\
\Leftrightarrow {\left( {x - 3} \right)^6} - {\left( {x - 3} \right)^3} = 0\\
\Leftrightarrow {\left( {x - 3} \right)^3}.\left[ {{{\left( {x - 3} \right)}^3} - 1} \right] = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 3 = 0\\
{\left( {x - 3} \right)^3} = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x - 3 = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = 4
\end{array} \right.\\
Vậy\,x = 3;x = 4
\end{array}$
