\(\left(x+1\right)^{2022}=\left(x+1\right)^2\)
\(\Rightarrow\left(x+1\right)^{2022}-\left(x+1\right)^2=0\)
\(\Rightarrow\left(x+1\right)^2\left[\left(x+1\right)^{2020}-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x+1\right)^2=0\\\left(x+1\right)^{2020}-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)
Vậy: ...
\(\left(x+1\right)^{2022}\text{=}\left(x+1\right)^2\)
\(\left(x+1\right)^2.\left(x+1\right)^{1011}\text{=}\left(x+1\right)^2\)
\(\Rightarrow\left(x+1\right)^{1011}\text{=}1\)
\(x+1\text{=}1\)
x=0
Vậy.........
Có các trường hợp thoả mãn:
\(\left(x+1\right)=0\)
\(\left(x+1\right)=1\)
\(\left(x+1\right)=-1\)
Vậy giá trị của x lần lượt: \(-1;0;-2\)
\(\left(x+1\right)^{2022}=\left(x+1\right)^2\)
\(\Rightarrow\left(x+1\right)^{2022}-\left(x+1\right)^2=0\)
\(\Rightarrow\left(x+1\right)^2\left[\left(x+1\right)^{2020}-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^{2020}-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\\left(x+1\right)^{2020}=1=1^{2020}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x+1=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)
