Question

\(\dfrac{\left(x-3\right)^2+\left(y-2\right)^2=0\left(1\right)}{ }\) :((( giúp mik vớiiii

Answer 1

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

Answer 2

\(Ta\ thấy :\)

\(\left(x-3\right)^2\ge0\forall x;\left(y-2\right)^2\ge0\forall y\\ =>\left(x-3\right)^2+\left(y-2\right)^2\ge0\forall x;y\)

\(Dấu''=''\ xảy\ ra\ khi :\)

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

\(Vậy\ (x;y)=(3;2)\)