Question

(x+1)²+(y-3)²=0

Answer 1

Vì : \(\left(x+1\right)^2\ge0\forall x\)

\(\left(y-3\right)^2\ge0\forall y\)

\(\Rightarrow\left(x+1\right)^2+\left(y-3\right)^2\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\y-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\y=3\end{matrix}\right.\)

Vậy x = -1 ; y = 3

Answer 2

Vì \(\left\{{}\begin{matrix}\left(x+1\right)^2\ge0\forall x\\\left(y-3\right)^2\ge0\forall y\end{matrix}\right.\)

=> Để biểu thức có gt = 0 thì

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

Vậy ................

Answer 3

\(\left(x+1\right)^2+\left(y-3\right)^2=0\) => \(\left(x+1\right)^2=\left(y-3\right)^2\) ; Mà \(\left(x+2\right)^2\ge0\) ; \(\left(y-3\right)^2\ge0\) \(\Rightarrow\left[{}\begin{matrix}\left(x+1\right)^2=0\\\left(y-3\right)^2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+1=0\\y-3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\y=3\end{matrix}\right.\) Vậy x = -1 ; y =3