Question

( x -1 ) ² + ( 3 - y ) ² ≤ 0

Answer 1

Đề là gì bn nhỉ

Answer 2

\(\left(x-1\right)^2+\left(3-y\right)^2\text{≤}\) (1)

Vì \(\left\{{}\begin{matrix}\left(x-1\right)^2\text{≥}0\\\left(3-y\right)^2\text{≥}0\end{matrix}\right.\)

⇒\(\left(x-1\right)^2+\left(3-y\right)^2\text{≥}0\) (2)

Từ (1) và (2):

⇒\(\left(x-1\right)^2+\left(3-y\right)^2=0\)

⇒\(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(3-y\right)^2=0\end{matrix}\right.\) ⇒\(\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)