\(a,\left(x-3\right)^{27}=\left(x-3\right)^{127}\)
\(\Leftrightarrow\left(x-3\right)^{127}-\left(x-3\right)^{27}=0\)
\(\Leftrightarrow\left(x-3\right)^{27}\left[\left(x-3\right)^{100}-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-3\right)^{27}=0\\\left(x-3\right)^{100}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-3=0\\\left(x-3\right)^{100}=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x-3=\pm1\end{cases}\Rightarrow}x\in\left\{2;3;4\right\}}\)
Vậy \(x\in\left\{2;3;4\right\}\)
\(b,\left(x-1\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)
Vậy \(x\in\left\{1;2\right\}\)
A:suy ra x-3=1 hoặc x-3=0
x=4 hoặc x=3
B:
vì(x-1)(x-2)=0
suy ra x-1=0 hoặc x-2=0
x=1 hoặc x=2
