Ta co : \(x^2+y^2-4x+3=0\)
\(=>\left(x-2\right)^2+y^2=1\)
\(=>\left(x-2\right)^2\le1=>x\le3\)
Lai co : \(x^2+y^2=4x-3\le4.3-3=9\)
Dau = xay ra \(< =>\hept{\begin{cases}x=4\\y=0\end{cases}}\)
Vay gtln cua P = 9 khi x = 4 ; y = 0
(sai thi bo qua cho minh vi lan dau lam dang nay)