(1)若a^2+2a+b^2-6b+10=0,求a^2-b^2的值(2)若△ABC三边a,b,c满足a^2+b^2+c^2=ab+bc+ca判断△ABC的形状

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/26 17:29:33
(1)若a^2+2a+b^2-6b+10=0,求a^2-b^2的值(2)若△ABC三边a,b,c满足a^2+b^2+c^2=ab+bc+ca判断△ABC的形状
x͒J@F͐l@}% xh案7ETD E10b n6xPz?;_e1T&P 3[x,|C u'**[(MJzRK:*]˫-A+z tJ:H[Q6 403/E udb_,lt

(1)若a^2+2a+b^2-6b+10=0,求a^2-b^2的值(2)若△ABC三边a,b,c满足a^2+b^2+c^2=ab+bc+ca判断△ABC的形状
(1)若a^2+2a+b^2-6b+10=0,求a^2-b^2的值
(2)若△ABC三边a,b,c满足a^2+b^2+c^2=ab+bc+ca判断△ABC的形状

(1)若a^2+2a+b^2-6b+10=0,求a^2-b^2的值(2)若△ABC三边a,b,c满足a^2+b^2+c^2=ab+bc+ca判断△ABC的形状
(1)若a²+2a+b²-6b+10=0,求a^2-b^2的值
a²+2a+1+b²-6b=9=0
(a+1)²+(b-3)²=0
a=-1,b=3
a²-b²=(-1)²-(3)²=1-9=-8
(2)若△ABC三边a,b,c满足a^2+b^2+c^2=ab+bc+ca判断△ABC的形状
2a²+2b²+2c²=2ab+2ac+2bc
2a²+2b²+2c²-2ab-2ac-2bc=0
(a²-2ab+b²)+(a²-2ac+c²)+(b²-2bc+c²)=0
(a-b)²+(a-c)²+(b-c)²=0
(a-b)=(a-c)=(b-c)=0
a=b=c
△ABC是等边三角形

-8
等边三角形