对关于x的一元二次方程ax+bx+c=0(a≠0),如果a+b+c=0,那么它必有一个根为x1=1,另一个根为x2=c/a.证明:因为a+b+c=0,所以c=-a-b.把c=-a-b带入原方程,得ax+bx-a-b=0.即a(x-1)+b(x-1)=0.所以a(x+1)(x-1)+b(x-1)=0.所以(x-1)(ax
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![对关于x的一元二次方程ax+bx+c=0(a≠0),如果a+b+c=0,那么它必有一个根为x1=1,另一个根为x2=c/a.证明:因为a+b+c=0,所以c=-a-b.把c=-a-b带入原方程,得ax+bx-a-b=0.即a(x-1)+b(x-1)=0.所以a(x+1)(x-1)+b(x-1)=0.所以(x-1)(ax](/uploads/image/z/7180781-5-1.jpg?t=%E5%AF%B9%E5%85%B3%E4%BA%8Ex%E7%9A%84%E4%B8%80%E5%85%83%E4%BA%8C%E6%AC%A1%E6%96%B9%E7%A8%8Bax%2Bbx%2Bc%3D0%28a%E2%89%A00%29%2C%E5%A6%82%E6%9E%9Ca%2Bb%2Bc%3D0%2C%E9%82%A3%E4%B9%88%E5%AE%83%E5%BF%85%E6%9C%89%E4%B8%80%E4%B8%AA%E6%A0%B9%E4%B8%BAx1%3D1%2C%E5%8F%A6%E4%B8%80%E4%B8%AA%E6%A0%B9%E4%B8%BAx2%3Dc%2Fa.%E8%AF%81%E6%98%8E%3A%E5%9B%A0%E4%B8%BAa%2Bb%2Bc%3D0%2C%E6%89%80%E4%BB%A5c%3D-a-b.%E6%8A%8Ac%3D-a-b%E5%B8%A6%E5%85%A5%E5%8E%9F%E6%96%B9%E7%A8%8B%2C%E5%BE%97ax%2Bbx-a-b%3D0.%E5%8D%B3a%28x-1%29%2Bb%28x-1%29%3D0.%E6%89%80%E4%BB%A5a%28x%2B1%29%28x-1%29%2Bb%28x-1%29%3D0.%E6%89%80%E4%BB%A5%28x-1%29%28ax)
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