a、b、c是△ABC的三边,若方程2ax²+2\sqrt{b²+c²}x+(b+c)=a有两个相等的实数根若方程2ax²+2\sqrt{b²+c²}x+(b+c)=a有两个相等的实数根,判断△ABC的形状
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/04 01:12:15
![a、b、c是△ABC的三边,若方程2ax²+2\sqrt{b²+c²}x+(b+c)=a有两个相等的实数根若方程2ax²+2\sqrt{b²+c²}x+(b+c)=a有两个相等的实数根,判断△ABC的形状](/uploads/image/z/4112380-28-0.jpg?t=a%E3%80%81b%E3%80%81c%E6%98%AF%E2%96%B3ABC%E7%9A%84%E4%B8%89%E8%BE%B9%2C%E8%8B%A5%E6%96%B9%E7%A8%8B2ax%26%23178%3B%2B2%5Csqrt%7Bb%26%23178%3B%2Bc%26%23178%3B%7Dx%2B%28b%2Bc%29%3Da%E6%9C%89%E4%B8%A4%E4%B8%AA%E7%9B%B8%E7%AD%89%E7%9A%84%E5%AE%9E%E6%95%B0%E6%A0%B9%E8%8B%A5%E6%96%B9%E7%A8%8B2ax%26%23178%3B%2B2%5Csqrt%7Bb%26%23178%3B%2Bc%26%23178%3B%7Dx%2B%28b%2Bc%29%3Da%E6%9C%89%E4%B8%A4%E4%B8%AA%E7%9B%B8%E7%AD%89%E7%9A%84%E5%AE%9E%E6%95%B0%E6%A0%B9%2C%E5%88%A4%E6%96%AD%E2%96%B3ABC%E7%9A%84%E5%BD%A2%E7%8A%B6)
a、b、c是△ABC的三边,若方程2ax²+2\sqrt{b²+c²}x+(b+c)=a有两个相等的实数根若方程2ax²+2\sqrt{b²+c²}x+(b+c)=a有两个相等的实数根,判断△ABC的形状
a、b、c是△ABC的三边,若方程2ax²+2\sqrt{b²+c²}x+(b+c)=a有两个相等的实数根
若方程2ax²+2\sqrt{b²+c²}x+(b+c)=a有两个相等的实数根,判断△ABC的形状
a、b、c是△ABC的三边,若方程2ax²+2\sqrt{b²+c²}x+(b+c)=a有两个相等的实数根若方程2ax²+2\sqrt{b²+c²}x+(b+c)=a有两个相等的实数根,判断△ABC的形状
△=b²-4ac=0
4(b²+c²)-8a(b+c-a)=0
b²+c²-2ab-2ac+2a²=0
a²-2ab+b²+a²-2ac+c²=0
(a-b)²+(a-c)²=0
a-b=0 a-c=0
a=b=c
△ABC是等边三角形