Tìm số nguyên x,y biết :
\(\left(y+3\right)^2+\left(x+17\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\left(y+3\right)^2=0\\\left(x+17\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}y=-3\\x=-17\end{cases}}}\)
Vậy \(y=-3;x=-17\)
\(\left(y+3\right)^2+\left(x+17\right)^2=0\)
=>\(\left(y+3\right)^2=\left(x+17\right)^2=0\)
=>\(y+3=x+17=0\)
=>\(\hept{\begin{cases}y=\left(-3\right)\\x=\left(-17\right)\end{cases}}\)
Vậy x = -17; y = -3
(y+3)2+(x+17)2=0
Ta có: (y+3)2 >=0 với mọi y
(x+17)2 >=0 với mọi x
=> (y+3)2+(x+17)2 >=0
mà (y+3)2+(x+17)2=0
=> \(\hept{\begin{cases}\left(y+3\right)^2=0\\\left(x+17\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}y+3=0\\x+17=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y=-3\\x=-17\end{cases}}}\)
