*Lời giải :
`P = (x^2 - 2)^2 + (y - 1)^2`
Vì \(\left\{ \begin{array}{l}(x^2 - 2)^2≥0∀x\\(y-1)^2≥0∀y\end{array} \right.\)
`-> (x^2 - 2)^2 + (y -1)^2 ≥ 0∀x,y`
`-> P_{min} = 0`
Khi :
`⇔` \(\left\{ \begin{array}{l}x^2 - 2 = 0\\y-1=0\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}x=±\sqrt{2}\\y=1\end{array} \right.\)