Question

已知x≠y，且x²-x=13，y²-y=13，求x+y-7的值

Answer 1

换个思粗返维思考！
整理，x²-x-13=0，核燃y²-y-13=0，
又x≠改凳虚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

Answer 2

x²-x=13，y²-y=13
两式碧绝和相减悔盯得宏猛
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

Answer 3

x²森脊-x=13.....(1)
y²-y=13.....(2)
(1)-(2)得
(x-y)(x+y-1)=0,因x≠唯数y,x-y≠0,所以
x+y-1=0,两边同时加此山渗上-6，得
x+y-7=-6

Answer 4

x²-x=y²-y
x²山带-y²拿亮=x-y
(x+y)(x-y)=x-y
x≠消唯宽y
x+y=1
x+y-7=-6

Answer 5

x²-x=13，y²塌灶扰-y=13

∴x²团旦辩睁-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