已知x,y,z,均为正整数,且满足x²+z²=10,z²+y²=13,求(x-y)²的值

2025-05-14 15:45:45
推荐回答(3个)
回答1:

已知x,y,z,均为正整数,且满足x²+z²=10,z²+y²=13,
x²<10
z²<10
y²<13

x²+z²=10,
z²=10-x²

z²+y²=13

10-x²+y²=13
y²-x²=3
y=2 x=1
z=3
(x-y)²的值

=(1-2)²
=1

回答2:

∵z²+y²-(x²+z²)=3,

∴(y-x)(y+x)=3,
∵x,y,z是正整数,x+y≥2

∴y+x=3,所以y-x=1,解得y=2,x=1,所以(x-y)²=1.

回答3:

x=1,y=2,z=3,(x-y)(x-y)=1