已知方程(ac-bc)x^+(bc-ab)x+(ab-ac)=0有两个相同的实数根试着说明2/b=1/a+1/c 如果好的话给奖金我急用啊******一元二次方程有两相等实根,则△=0[b*(c-a)]^2-4*a(b-c)*c(a-b)=0化简(bc)^2+(ba)^2+2acb^2-4ac(ba-ac+bc
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/01 14:12:44
![已知方程(ac-bc)x^+(bc-ab)x+(ab-ac)=0有两个相同的实数根试着说明2/b=1/a+1/c 如果好的话给奖金我急用啊******一元二次方程有两相等实根,则△=0[b*(c-a)]^2-4*a(b-c)*c(a-b)=0化简(bc)^2+(ba)^2+2acb^2-4ac(ba-ac+bc](/uploads/image/z/7188354-18-4.jpg?t=%E5%B7%B2%E7%9F%A5%E6%96%B9%E7%A8%8B%EF%BC%88ac-bc%29x%5E%2B%28bc-ab%29x%2B%28ab-ac%29%3D0%E6%9C%89%E4%B8%A4%E4%B8%AA%E7%9B%B8%E5%90%8C%E7%9A%84%E5%AE%9E%E6%95%B0%E6%A0%B9%E8%AF%95%E7%9D%80%E8%AF%B4%E6%98%8E2%2Fb%3D1%2Fa%2B1%2Fc+%E5%A6%82%E6%9E%9C%E5%A5%BD%E7%9A%84%E8%AF%9D%E7%BB%99%E5%A5%96%E9%87%91%E6%88%91%E6%80%A5%E7%94%A8%E5%95%8A%2A%2A%2A%2A%2A%2A%E4%B8%80%E5%85%83%E4%BA%8C%E6%AC%A1%E6%96%B9%E7%A8%8B%E6%9C%89%E4%B8%A4%E7%9B%B8%E7%AD%89%E5%AE%9E%E6%A0%B9%EF%BC%8C%E5%88%99%E2%96%B3%3D0%5Bb%2A%28c-a%29%5D%5E2-4%2Aa%28b-c%29%2Ac%28a-b%29%3D0%E5%8C%96%E7%AE%80%28bc%29%5E2%2B%28ba%29%5E2%2B2acb%5E2-4ac%28ba-ac%2Bbc)
xSr@}\&x Qö3{tXK (u
ZZZ8Rx+^/qttBfB}~9{ZZ>ٛ"&ZaSW11
ϰr]~Fwvh5?
kGU=ke"2B u8tM֬tf6+)=_*{w.u/Njz