Đáp án:
\(\left[ \begin{array}{l}
x = - 1\\
x = 0\\
x = - 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
b){\left( {x + 1} \right)^{x + 2}} - {\left( {x + 1} \right)^{x + 10}} = 0\\
\to {\left( {x + 1} \right)^x}.{\left( {x + 1} \right)^2} - {\left( {x + 1} \right)^x}.{\left( {x + 1} \right)^{10}} = 0\\
\to {\left( {x + 1} \right)^x}.{\left( {x + 1} \right)^2}\left( {1 - {{\left( {x + 1} \right)}^8}} \right) = 0\\
\to \left[ \begin{array}{l}
{\left( {x + 1} \right)^x} = 0\\
{\left( {x + 1} \right)^2} = 0\\
1 - {\left( {x + 1} \right)^8} = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x + 1 = 0\\
{\left( {x + 1} \right)^8} = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\\
x + 1 = 1\\
x + 1 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\\
x = 0\\
x = - 2
\end{array} \right.
\end{array}\)
