Ta có :
\(\left(x-10\right)^{x+1}-\left(x-10\right)^{x+11}=0\)
\(\Leftrightarrow\left(x-10\right)^{x+1}\left[1-\left(x-10\right)^{10}\right]=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-10=0\\1-\left(x-10\right)^{10}=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=10\\\left[\begin{array}{nghiempt}x-10=1\\x-10=-1\end{array}\right.\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=10\\\left[\begin{array}{nghiempt}x=11\\x=9\end{array}\right.\end{array}\right.\)
Vậy x = 10 ; x = 11 ; x = 9
\(\left(x-10\right)^{x+1}-\left(x-10\right)^{x+11}=0\)
\(\Rightarrow\left(x-10\right)^{x+1}.\left[1-\left(x-10\right)^{10}\right]=0\)
\(\Rightarrow\left(x-10\right)^{x+1}=0\) hoặc \(1-\left(x-10\right)^{10}=0\)
+) \(\left(x-10\right)^{x+1}=0\)
\(\Rightarrow x-10=0\)
\(\Rightarrow x=10\)
+) \(1-\left(x-10\right)^{10}=0\)
\(\Rightarrow\left(x-10\right)^{10}=1\)
\(\Rightarrow x-10=\pm1\)
+ \(x-10=1\Rightarrow x=11\)
+ \(x-10=-1\Rightarrow x=9\)
Vậy \(x\in\left\{10;11;9\right\}\)
