Question

Tìm các số nguyên x,y sao cho :

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

b)\(/2x-6/+/y+7/=0\)

Answer 1

ta có \(\hept{\begin{cases}\left(x-2\right)^2\ge0\\\left(y+1\right)^2\ge0\end{cases}}\)

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

\(\Rightarrow\hept{\begin{cases}x-2=0\\y+1=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=2\\y=-1\end{cases}}\)

b)

ta có  \(\hept{\begin{cases}\left|2x-6\right|\ge0\\\left|y+7\right|\ge0\end{cases}}\)

mà \(\left|2x-6\right|+\left|y+7\right|=0\)

\(\Rightarrow\hept{\begin{cases}2x-6=0\\y+7=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\y=-7\end{cases}}}\)

Answer 2

tốt lắm

Answer 3

tốt j bạn

Answer 4

bài quá dễ