已知x≠y,且x²-x=13,y²-y=13,求x+y-7的值

已知x≠y,且x²-x=13,y²-y=13,求x+y-7的值
已知x≠y,且x²-x=13,y²-y=13,求x+y-7的值

已知x≠y,且x²-x=13,y²-y=13,求x+y-7的值
x²-x=13,y²-y=13
∴x²-x-(y²-y)=13-13
x²-y²-x+y=0
(x+y)(x-y)-(x-y)=0
(x-y)(x+y-1)=0
∴x+y-1=0
x+y=1
x+y-7=1-7=-6

x²-x=y²-y
x²-y²-x+y=0
(x+y)(x-y)-(x-y)=0
(x-y)(x+y-1)=0
x<>y, 所以x+y-1=0, x+y=1
x+y-7=1-7=-6

x²-x=y²-y
x²-y²=x-y
(x+y)(x-y)=x-y
x≠y
x+y=1
x+y-7=-6

换个思维思考！
整理，x²-x-13=0，y²-y-13=0，
又x≠y
所以x,y是方程t²-t-13=0的两根
由根与系数关系，得，
x+y=1
所以x+y-7=1-7=-6

x²-x=13，y²-y=13
两式相减得
x^2-y^2-x+y=0
(x+y)(x-y)-(x-y)=0
(x-y)(x+y-1)=0
x+y-1=0
x+y=1
x+y-7=-6

x²-x=13.....(1)
y²-y=13.....(2)
(1)-(2)得
(x-y)(x+y-1)=0,因x≠y,x-y≠0,所以
x+y-1=0,两边同时加上-6，得
x+y-7=-6