数学问题已知正整数x,y满足x^2-y^2=24,求xy(x+y)的值
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数学问题已知正整数x,y满足x^2-y^2=24,求xy(x+y)的值
数学问题已知正整数x,y满足x^2-y^2=24,求xy(x+y)的值
数学问题已知正整数x,y满足x^2-y^2=24,求xy(x+y)的值
x^2-y^2=24
(x+y)(x-y)=24
24=1×24=2×12=3×8=4×6
x,y为正整数,x,y可能满足的方程如下:
x+y=24
x-y=1
解得x,y不为整数,舍去.
x+y=12
x-y=2
解得x=7 y=5
x+y=8
x-y=3
解得x,y不为整数,舍去
x+y=6
x-y=4
解得x=5 y=1
x=7 y=5或x=5 y=1
xy(x+y)
=7×5×(7+5)
=420
或
xy(x+y)
=5×1×(5+1)
=30
xy(x+y)的值为420或30
x^2-y^2=24
(x-y)*(x+y)=24
又因为x, y是正整数,所以x-y为正,所以两种可能:
1: x=7, y=5
2: x=5, y=1
所以
1:xy(x+y)=35*12=420
2: xy(x+y)=5*6=30