Question

Answer 1

\(\left(x+1\right)^{x+10}=\left(x+1\right)^{x+4}\)

\(\Rightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\)

\(\Rightarrow\left(x+1\right)^{x+4}\left[\left(x+1\right)^{x+6}+1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x+1\right)^{x+4}=0\\\left(x+1\right)^{x+6}-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^{x+6}=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x+1=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)

Answer 2

\(\left(x+1\right)^{x+10}=\left(x+1\right)^{x+4}\)

\(\Rightarrow\left(x+1\right)^x.\left(x+1\right)^{10}=\left(x+1\right)^x.\left(x+1\right)^4\)

\(\Rightarrow\left(x+1\right)^{10}=\left(x+1\right)^4\)

\(\Rightarrow\left(x+1\right)^6=1\\ \Rightarrow\left[{}\begin{matrix}x+1=-1\\x+1=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-2\\x=0\end{matrix}\right.\)

Answer 3

Ta có: \(\left(x+1\right)^{x+10}=\left(x+1\right)^{x+4}\)

\(\Leftrightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\)

\(\Leftrightarrow\left(x+1\right)^{x+4}\cdot\left[\left(x+1\right)^6-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\\x+1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\)