xy-2x-2y+8=0;
(x-2)(y-2)+4=0;
(x-2)(y-2)=-4=(-2)x2.
x≤y,則x-2≤y-2,即x-2=-2;y-2=2.
得x=0,y=4.
xy-2x-2y+8=0的整數(shù)解(x≤y)為
xy-2x-2y+8=0的整數(shù)解(x≤y)為
數(shù)學(xué)人氣：484 ℃時(shí)間：2020-06-21 11:20:33
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