若a b c为实数且a²+b²+c²-ab-bc-ca=0求证a=b=c急求
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/30 05:03:08
![若a b c为实数且a²+b²+c²-ab-bc-ca=0求证a=b=c急求](/uploads/image/z/1143879-15-9.jpg?t=%E8%8B%A5a+b+c%E4%B8%BA%E5%AE%9E%E6%95%B0%E4%B8%94a%26%23178%3B%2Bb%26%23178%3B%2Bc%26%23178%3B-ab-bc-ca%3D0%E6%B1%82%E8%AF%81a%3Db%3Dc%E6%80%A5%E6%B1%82)
若a b c为实数且a²+b²+c²-ab-bc-ca=0求证a=b=c急求
若a b c为实数且a²+b²+c²-ab-bc-ca=0求证a=b=c急求
若a b c为实数且a²+b²+c²-ab-bc-ca=0求证a=b=c急求
a²+b²+c²-ab-bc-ca=0
两边同乘以2得:
2a²+2b²+2c²-2ab-2bc-2ca=0
(a-b)²+(b-c)²+(a-c)²=0
则:必有:
a-b=b-c=a-c=0
解得:a=b=c
证明:∵a²+b²+c²-ab-bc-ca=0
∴2a²+2b²+2c²-2ab-2bc-2ca=0
(a²-2ab+b²)+(a²-2ac+c²)+(b²-2bc+c²)=0
(a-b)²+(a-c)²+(b-c)²=0
∴(a-b)²=0, (a-c)²=0, (b-c)²=0
a-b=0, a-c=0, b-c=0
∴a=b, a=c, b=c
即:a=b=c.
a²+b²+c²-ab-ac-bc=0
2a²+2b²+2c²-2ab-2ac-2bc=0
﹙a²-2ab+b²﹚+﹙b²-2bc+c²﹚+﹙a²-2ac+c²﹚=0
﹙a-b﹚²+﹙b-c﹚²+﹙a-c﹚...
全部展开
a²+b²+c²-ab-ac-bc=0
2a²+2b²+2c²-2ab-2ac-2bc=0
﹙a²-2ab+b²﹚+﹙b²-2bc+c²﹚+﹙a²-2ac+c²﹚=0
﹙a-b﹚²+﹙b-c﹚²+﹙a-c﹚²=0
∵ ﹙a-b﹚²≥0, ﹙b-c﹚²≥0, ﹙a-c﹚²≥0
∴ a-b=0, a=b
b-c=0, b=c
a-c=0, a=c
∴ a=b=c. 证毕!
收起