1, ta có x2+y2=0
do \(x^2\ge0,y^2\ge0\Rightarrow x^2+y^2\ge\)0
mà x2+y2=0
=>x2=0,y2=0
=> x=y=0
Phần 2 tương tự nha bn
3,|x-3|+|y+2|=0
do\(\left|x-3\right|\ge0,\left|y+2\right|\ge0\)
\(\Rightarrow\left|x-3\right|+\left|y+2\right|\ge0\)
\(\Rightarrow\hept{\begin{cases}\left|x-3\right|=0\\\left|y+2\right|=0\end{cases}\Rightarrow}\hept{\begin{cases}x-3=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\y=-2\end{cases}}}\)
Phần 4 đề thiếu nhé bn
1. \(x^2+y^2=0\)
\(\Rightarrow\hept{\begin{cases}x^2=0\\y^2=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0\\y=0\end{cases}}\)
2.\(|x|+|y|=0\)
\(\Rightarrow\hept{\begin{cases}|x|=0\\|y|=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0\\y=0\end{cases}}\)
3.\(|x-3|+|y+2|=0\)
\(\Rightarrow\hept{\begin{cases}|x-3|=0\\|y+2|=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x-3=0\\y+2=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=3\\y=-2\end{cases}}\)
4. ý 4 bạn ghi thiếu đề bài rồi
\(\left(1\right)x^2+y^2=0\)
Vì \(x^2,y^2\ge0\)
\(\Rightarrow x^2=y^2=0\)
\(\Rightarrow x=y=0\)
\(\left(2\right)|x|+|y|=0\)
Mà \(|x|,|y|\ge0\)
\(\Rightarrow|x|=|y|=0\)
\(\Rightarrow x=y=0\)
\(\left(3\right)|x-3|+|y+2|=0\)
Mà \(|x-3|,|y+2|\ge0\)
\(\Rightarrow\hept{\begin{cases}|x-3|=0\\|y+2|=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x-3=0\\y+2=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=3\\y=-2\end{cases}}\)
\(\left(4\right)\left(x+1\right)^2+\left(y+3\right)=??????\)
