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(x-7)x+1-(x-7)x+11=0
<=> (x-7)x+1=(x-7)x+11
<=> x+1=x+11
<=> x-x=11-1
<=> 0x=10
=> PT vô nghiệm
\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
\(\Leftrightarrow x+1=x+11\) (vô lý)
TH1: \(x-7=1\Rightarrow x=8\)
TH2: \(x-7=-1\Rightarrow x=6\)
TH3: \(x-7=0\Rightarrow x=7\)
Vậy:....
(x-7)x+1( 1 - (x-7)10) = 0
(x-7)x+1 = 0
x = 7
1- (x-7)10 = 0
(x - 7)10 = 1
x =8
\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
\(\Leftrightarrow\left(x-7\right)^{x+1}.\left[1-\left(x-7\right)^{x+10}\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+10}=0\\1-\left(x-7\right)^{x+10}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\\left(x-7\right)^{x+10}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\\left[{}\begin{matrix}x-7=1\\x-7=-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\\left[{}\begin{matrix}x=8\\x=6\end{matrix}\right.\end{matrix}\right.\)
Vậy ....
\(\left(x-7\right)^{x-1}-\left(x-7\right)^{x+11}=0\)
\(\Rightarrow\left(x-7\right)^{x+1}.\left[1-\left(x-7\right)^{10}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{10}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x-7=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0+7\\x-7=1\\x-7=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=7\\x=1+7\\x=\left(-1\right)+7\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)
Vậy \(x\in\left\{7;8;6\right\}.\)
Chúc bạn học tốt!