Ответ:
х^2+у^2≤16.
x>/= 2-y.
→(2-y)^2+y^2<\=16. 4-y^2+y^2-16</=0.
4-16=4
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Ответ:
х^2+у^2≤16.
x>/= 2-y.
→(2-y)^2+y^2<\=16. 4-y^2+y^2-16</=0.
4-16=4