\(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left(x^2-2x\right)=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left(x-2\right)x=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\left(x-1\right)^{x+2}=0\\x-2=0\\x=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2\\x=0\end{array}\right.\)
Ta có: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Rightarrow x+4=x+2\)\(\in\varnothing\)
\(\Rightarrow\) vô nghiệm
Chắc sai ak 
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