如果x1,x2是一元二次方程ax²+bx+c=0的两根,那么有x1+x2=-b/a,x1x2=c/a.已知α,β是方程x²+(m-2)x+1=0的两个根,求(1+mα+α²)×(1+mβ+β²)的值.
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![如果x1,x2是一元二次方程ax²+bx+c=0的两根,那么有x1+x2=-b/a,x1x2=c/a.已知α,β是方程x²+(m-2)x+1=0的两个根,求(1+mα+α²)×(1+mβ+β²)的值.](/uploads/image/z/1982544-24-4.jpg?t=%E5%A6%82%E6%9E%9Cx1%2Cx2%E6%98%AF%E4%B8%80%E5%85%83%E4%BA%8C%E6%AC%A1%E6%96%B9%E7%A8%8Bax%26%23178%3B%EF%BC%8Bbx%EF%BC%8Bc%3D0%E7%9A%84%E4%B8%A4%E6%A0%B9%2C%E9%82%A3%E4%B9%88%E6%9C%89x1%EF%BC%8Bx2%3D%EF%BC%8Db%2Fa%2Cx1x2%3Dc%2Fa.%E5%B7%B2%E7%9F%A5%CE%B1%2C%CE%B2%E6%98%AF%E6%96%B9%E7%A8%8Bx%26%23178%3B%EF%BC%8B%EF%BC%88m%EF%BC%8D2%EF%BC%89x%EF%BC%8B1%3D0%E7%9A%84%E4%B8%A4%E4%B8%AA%E6%A0%B9%2C%E6%B1%82%EF%BC%881%EF%BC%8Bm%CE%B1%EF%BC%8B%CE%B1%26%23178%3B%EF%BC%89%C3%97%EF%BC%881%EF%BC%8Bm%CE%B2%EF%BC%8B%CE%B2%26%23178%3B%EF%BC%89%E7%9A%84%E5%80%BC.)
如果x1,x2是一元二次方程ax²+bx+c=0的两根,那么有x1+x2=-b/a,x1x2=c/a.已知α,β是方程x²+(m-2)x+1=0的两个根,求(1+mα+α²)×(1+mβ+β²)的值.
如果x1,x2是一元二次方程ax²+bx+c=0的两根,那么有x1+x2=-b/a,x1x2=c/a.已知α,β是方程x²+(m-2)x+1=0的两个根,求(1+mα+α²)×(1+mβ+β²)的值.
如果x1,x2是一元二次方程ax²+bx+c=0的两根,那么有x1+x2=-b/a,x1x2=c/a.已知α,β是方程x²+(m-2)x+1=0的两个根,求(1+mα+α²)×(1+mβ+β²)的值.
由韦达定理,得
α+β=2-m
αβ=1
(1+mα+α²)×(1+mβ+β²)
=(αβ+mα+α²)×(αβ+mβ+β²)
=α(β+m+α)×β(α+m+β)
=αβ*(α+m+β)²
=1*(2-m+m)²
=2²
=4
α,β是方程x²+(m-2)x+1=0的两个根,αβ=1,α+β=2-m
(1+mα+α²)×(1+mβ+β²)=(αβ+mα+α²)×(αβ+mβ+β²)=α(α+β+m)β(α+β+m)=αβ(α+β+m)²=(α+β+m)²=(2-m+m)²=2²=4