Question

Tìm x biết

\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\) và x \(\in\) Z

Answer 1

\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)

\(\Leftrightarrow\left(x-1\right)^{x+2}-\left(x-1\right)^{x+6}=0\)

\(\Leftrightarrow\left(x-1\right)^{x+2}\left[1-\left(x-1\right)^4\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^4=0\end{matrix}\right.\)

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

Vậy..............

Answer 2

(x-1)=±1 la nghiem x{0;2} la nghiem

x-1=0 la nghiem => x =1LA NGHIEM

X≠{0,1,2}

chia hai ve cho (x-1)^{x+2}

<=>1=(x-1)^4

|x-1|=1

x={0,2}

kl

x€{0,1,2}