已知x和y都是正整数,并且满足条件中xy+x+y=71.x^2y+xy^2=880.求3x^2+8xy+3y^2的值

2025-06-27 11:52:27
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答:正整数x和y:xy+x+y=71x^2y+xy^2=880xy(x+y)=880因为:xy=71-(x+y)所以:(x+y)*[71-(x+y)]=880所以:-(x+y)^2+71(x+y)=880所以:(x+y)^2-71(x+y)+880=0所以:(x+y-55)(x+y-16)=0所以:x+y=55或者x+y=161)x+y=55...