Question

已知x+y=3,xy=-10,求（1）x^2-xy=y^2;(2）｜x-y ｜

Answer 1

已知x+y=3,xy=-10,求（1）x²-xy+y²; (2）｜x-y ｜
解一：∵x+y=3, xy=-10,∴x，y是方程z²-3z-10=(z-5)(z+2)=0的根：z₁=x=-2, z₂=y=5
或z₁=y=-2; z₂=x=5. ∴│x-y│=│-2-5│=│5+2│=7
(1)x²-xy=y²可能有误，应是x²-xy+y²或x²-xy-y²
x²-xy+y²=4+10+25=39; x²-xy-y²=4+10-25=11或25+10-4=31.
解二：
(x+y)²=x²+2xy+y²=x²-20+y²=9
故x²+y²=29 , x²-xy+y²=29-(-10)=39
(x-y)²=x²-2xy+y²=29+20=49
故│x-y│=7

Answer 2

1、
(x+y)^2=3^2
x^2+2xy+y^2=9
两边减去3xy=-30
x^2-xy+y^2=9+30=39

2、
(x-y)^2
=x^2-2xy+y^2
=(x^2-xy+y^2)-xy
=39+10
=49
所以|x-y|=7