aX²+bX+c+bX²+cX+a+cX²aX+baX²+bX+c=0,bX²+cX+a=0 cX²aX+b=0,有公共根,求证a+b+c=0
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![aX²+bX+c+bX²+cX+a+cX²aX+baX²+bX+c=0,bX²+cX+a=0 cX²aX+b=0,有公共根,求证a+b+c=0](/uploads/image/z/1624487-23-7.jpg?t=aX%26%23178%3B%2BbX%2Bc%2BbX%26%23178%3B%2BcX%2Ba%2BcX%26%23178%3BaX%2BbaX%26%23178%3B%2BbX%2Bc%3D0%EF%BC%8CbX%26%23178%3B%2BcX%2Ba%3D0+cX%26%23178%3BaX%2Bb%3D0%EF%BC%8C%E6%9C%89%E5%85%AC%E5%85%B1%E6%A0%B9%EF%BC%8C%E6%B1%82%E8%AF%81a%2Bb%2Bc%3D0)
aX²+bX+c+bX²+cX+a+cX²aX+baX²+bX+c=0,bX²+cX+a=0 cX²aX+b=0,有公共根,求证a+b+c=0
aX²+bX+c+bX²+cX+a+cX²aX+b
aX²+bX+c=0,bX²+cX+a=0 cX²aX+b=0,有公共根,求证a+b+c=0
aX²+bX+c+bX²+cX+a+cX²aX+baX²+bX+c=0,bX²+cX+a=0 cX²aX+b=0,有公共根,求证a+b+c=0
(a+b+c)X2+(a+b+c)X+(a+b+c)
=(a+b+c)(X2+x+1)
求值求根就随便你了
简单啊
X2+x+1=(X2+x+1/4)+3/4=(x+1/2)2+3/4>0
所以当且仅当a+b+c=0