已知x+y=3,x^2+y^2=7,求代数式(x-1)(y-1)的值怎么做?
问题描述：
已知x+y=3,x^2+y^2=7,求代数式(x-1)(y-1)的值怎么做?
答
∵x+y=3
∴x²+2xy+y²=9
∵x²+y²=7
∴2xy=9-7=2
∴xy=1
∴(x-1)(y-1)=xy-(x+y)+1=1-3+1=-1
答
x+y=3，x²+y²=7
(x+y)²=9 => x²+2xy+y²=9 => xy=1
(x-1)(y-1)=xy-(x+y)+1=1-3+1=-1
答
2xy=(x+y)^2-(x^2+y^2)=9-7=2
xy=1
(x-1)(y-1)
=xy-(x+y)+1
=1-3+1
=-1