\(\left(2x+1\right)^4=\left(2x+1\right)^6\)
Đặt 2x + 1 = a, ta có
\(a^4=a^6\)
\(\Rightarrow a^4-a^6=0\)
\(\Rightarrow a^4\left(1-a^2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a^4=0\\1-a^2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a=0\\a^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a=0\\a=1\\a=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\2x+1=1\\2x+1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=-1\\2x=0\\2x=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=0\\x=-1\end{matrix}\right.\)
\(\left(2x+1\right)^4=\left(2x+1\right)^6\)
\(\Rightarrow\left(2x+1\right)^4-\left(2x+1\right)^6=0\)
\(\Rightarrow\left(2x+1\right)^4.\left[\left(2x+1\right)^2-1\right]0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x+1\right)^4=0\\\left[\left(2x+1\right)^2-1\right]=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x+1=0\\\left(2x+1\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=0\end{matrix}\right.\)
\(\Rightarrow\left\{x_1=\dfrac{-1}{2};x_2=0\right\}\)
