Đề có cho x,y nguyên ko bạn nếu ko thì có nhìu x;y
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Vi \(\left(x-1\right)^2\) co so mu chan \(\Rightarrow\left(x-1\right)^2\ge0\) va \(\left(x-1\right)^2\) la so chinh phuong
Vi \(\left(y-1\right)^2\) co so mu chan \(\Rightarrow\left(y-1\right)^2\ge0\) va \(\left(y-1\right)^2\) la so chinh phuong
Ma \(\left(x-1\right)^2+\left(y-1\right)^2=4\)
Nen ta co bang sau:
\(\left(x-1\right)^2\) | 0 | 1 | 4 |
\(\left(y-1\right)^2\) | 4 | 3 | 0 |
x | 1 | 2 | 3 |
y | 3 | khong co | 1 |
Vay \(\left(x;y\right)\in\left\{\left(1;3\right);\left(3;1\right)\right\}\)
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