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

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...