已知x≠y,x²-x=5,y²-y=5,求x+y+7的值
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![已知x≠y,x²-x=5,y²-y=5,求x+y+7的值](/uploads/image/z/1101445-61-5.jpg?t=%E5%B7%B2%E7%9F%A5x%E2%89%A0y%2Cx%26%23178%3B-x%3D5%2Cy%26%23178%3B-y%3D5%2C%E6%B1%82x%2By%2B7%E7%9A%84%E5%80%BC)
已知x≠y,x²-x=5,y²-y=5,求x+y+7的值
已知x≠y,x²-x=5,y²-y=5,求x+y+7的值
已知x≠y,x²-x=5,y²-y=5,求x+y+7的值
∵x²-x=5,y²-y=0
∴(x²-x)-(y²-y)=0
x²-x-y²+y=0
(x²-y²)-(x-y)=0
(x+y)(x-y)-(x-y)=0
(x+y-1)(x-y)=0
∵x≠y
∴x-y≠0
∴x+y-1=0
即x+y=1
∴x+y+7
=(x+y)+7
=1+7
=8
(x²-x)-(y²-y)=0
x²-y²-x+y=0
(x+y)(x-y)-(x-y)=0
(x-y)(x+y-1)=0
∵x≠y
∴x+y-1=0
∴x+y=1
∴x+y+7=1+7=8
设x、y分别是方程t²-t-5=0的两个根。
由韦达定理知:x+y=1
∴ x+y+7=1+7=8.