Question

(x-3) mũ 2 + (x-y+2.5) mũ 4=0(x+1,5)2+(y−2,5)2≥0∀x,y

Answer 1

\(\left(x-3\right)^2>=0\forall x\)

\(\left(x-y+2,5\right)^4>=0\forall x,y\)

Do đó: \(\left(x-3\right)^2+\left(x-y+2,5\right)^4>=0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-3=0\\x-y+2,5=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=3\\y=x+2,5=3+2,5=5,5\end{matrix}\right.\)