\(\left(x-19\right)^{x+2000}-\left(x-19\right)^{x+2018}=0\) \(\rightarrow\left(x-19\right)^{x+2000}-\left(x-19\right)^{x+2000+18}=0\) \(\left(x-19\right)^{x+2000}-\left(x-19\right)^{x+2000}.\left(x-19\right)^{18}=0\) \(\left(x-19\right)^{x+2000}.\left[1-\left(x-19\right)^{18}\right]=0\) \(\Rightarrow\) \(\left\{{}\begin{matrix}\left(x-19\right)^{x+2000}=0\\1-\left(x-19\right)^{18}=0\end{matrix}\right.\) TH1 : \(\left(x-19\right)^{x+2000}=0\) \(\Leftrightarrow x-19=0\Rightarrow x=19\) TH2: \(\left(x-19\right)^{18}=0\) \(\Leftrightarrow\left(x-19\right)^{18}=1=1^{18}hoặc\left(-1\right)^{18}\) \(\Rightarrow\left[{}\begin{matrix}x-19=1\\x-19=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=20\\x=18\end{matrix}\right.\) Vậy \(x\in\left\{18;19;20\right\}\)
Để (x - 19)x+2000 - (x - 19)x+2018 = 0
Thì (x - 19)x+2000 = 0
=> x - 19 = 0
=> x = 19 (1)
Thì (x - 19)x+2018 = 0
=> x - 19 = 0
=> x = 19 (2)
Từ (1) và (2) suy ra x = 19
Thì (x - 19)x+2000 - (x - 19)x+2018 = 0
CHÚC BẠN HỌC TỐT!
Để : \(\left(x-19\right)^{x+2000}-\left(x-19\right)^{x+2018}=0\)
Thì : \(\left(x-19\right)^{x+2000}=\left(x-19\right)^{x+2018}\)
Hay : \(\left(x-19\right)^{x+2018}.\left(x-19\right)^2=\left(x-19\right)^{x+2018}\)
Do đó : \(\left(x-19\right)^2=1\)
\(\Leftrightarrow x-19=1\)
\(\Rightarrow x=20.\)
Vây \(x=20\)
