f(x-2)=x2-2x+3
=(x-2)2-2(x-2)+3
=x2-4x+4-2x+4-3
=x2-6x+5
=x2-x-5x+5
=x(x-1)-5(x-1)
=(x-5)(x-1)
=\(\left[{}\begin{matrix}x-5=0\\x-1=0\end{matrix}\right.\)
=\(\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
Đặt \(x-2=t\Rightarrow x=t+2\)
\(\Rightarrow f\left(t\right)=\left(t+2\right)^2-2\left(t+2\right)+3\)
\(\Rightarrow f\left(t\right)=t^2+2t+3\)
\(\Rightarrow f\left(x\right)=x^2+2x+3\)