\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)
\(\Leftrightarrow\left(x-1\right)^{x+6}-\left(x-1\right)^{x+2}=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}.\left[\left(x-1\right)^4-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^4-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x-1\right)^4=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
Vậy: x = 1 ; x = 2 hoặc x = 0.
Ta có: \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(x-1\right)^x-\left(x-1\right)^6\cdot\left(x-1\right)^x=0\)
\(\Leftrightarrow\left(x-1\right)^x\cdot\left[\left(x-1\right)^2-\left(x-1\right)^6\right]=0\)
\(\Leftrightarrow\left(x-1\right)^x\cdot\left(x-1\right)^2\cdot\left[1-\left(x-1\right)^4\right]=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\cdot\left[1-\left(x-1\right)^4\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^4=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x-1\in\left\{1;-1\right\}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x\in\left\{2;0\right\}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;1;2\right\}\)
\(\left(x-1\right)^x+2=\left(x-1\right)^x+6\)
\(-1^x+2=-1^x+6\)
\(-1x^x+2+1^x-6=0\)
\(-4\ne0\)vô nghiệm
(x - 1)x + 2 = (x - 1)x + 6 (x \(\in\) Z)
\(\Rightarrow\) (x - 1)x - (x - 1)x = 6 - 2
\(\Rightarrow\) 0 = 4
Vậy pt vô nghiệm (0 \(\ne\) 4)
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