求解法在△ABC中,若a^4+b^4+c^4-2a²c²-2b²c²+a²b²=0,则∠C=在△ABC中,若a^4+b^4+c^4-2a²c²-2b²c²+a²b²=0,则∠C=在△ABC中,若c²=a²+b²﹢ab,则
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![求解法在△ABC中,若a^4+b^4+c^4-2a²c²-2b²c²+a²b²=0,则∠C=在△ABC中,若a^4+b^4+c^4-2a²c²-2b²c²+a²b²=0,则∠C=在△ABC中,若c²=a²+b²﹢ab,则](/uploads/image/z/5185243-19-3.jpg?t=%E6%B1%82%E8%A7%A3%E6%B3%95%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2C%E8%8B%A5a%5E4%EF%BC%8Bb%5E4%EF%BC%8Bc%5E4%EF%BC%8D2a%26%23178%3Bc%26%23178%3B%EF%BC%8D2b%26%23178%3Bc%26%23178%3B%EF%BC%8Ba%26%23178%3Bb%26%23178%3B%3D0%2C%E5%88%99%E2%88%A0C%3D%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2C%E8%8B%A5a%5E4%EF%BC%8Bb%5E4%EF%BC%8Bc%5E4%EF%BC%8D2a%26%23178%3Bc%26%23178%3B%EF%BC%8D2b%26%23178%3Bc%26%23178%3B%EF%BC%8Ba%26%23178%3Bb%26%23178%3B%3D0%2C%E5%88%99%E2%88%A0C%3D%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2C%E8%8B%A5c%26%23178%3B%3Da%26%23178%3B%EF%BC%8Bb%26%23178%3B%EF%B9%A2ab%2C%E5%88%99)
求解法在△ABC中,若a^4+b^4+c^4-2a²c²-2b²c²+a²b²=0,则∠C=在△ABC中,若a^4+b^4+c^4-2a²c²-2b²c²+a²b²=0,则∠C=在△ABC中,若c²=a²+b²﹢ab,则
求解法在△ABC中,若a^4+b^4+c^4-2a²c²-2b²c²+a²b²=0,则∠C=
在△ABC中,若a^4+b^4+c^4-2a²c²-2b²c²+a²b²=0,则∠C=
在△ABC中,若c²=a²+b²﹢ab,则∠C=
求解法在△ABC中,若a^4+b^4+c^4-2a²c²-2b²c²+a²b²=0,则∠C=在△ABC中,若a^4+b^4+c^4-2a²c²-2b²c²+a²b²=0,则∠C=在△ABC中,若c²=a²+b²﹢ab,则
a^4+b^4+c^4-2a^2c^2-2b^2c^2+a^2b^2=0
a^4+b^4+c^4-2a^2c^2-2b^2c^2+2a^2b^2=a^2b^2
(c^2-a^2-b^2)^2=a^2b^2
c^2-a^2-b^2=±ab
根据余弦定理得:cosC=±1/2
C=60度 或120度
∠C=60°
∠C= 120°
a^4+b^4+c^4+2a^2b^2-2a^2c^2-2b^2c^2=a^2b^2
∵cosc=(a^2+b^2-c^2)/(2ab)
∴(cosc)^2=(a^4+b^4+c^4+2a^2b^2-2a^2c^2-2b^2c^2)/(4a^2b^2)
=(a^2b^2)/(4a^2b^2)
=1/4
∴cosc=±1/2.
所以∠C等于...
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a^4+b^4+c^4+2a^2b^2-2a^2c^2-2b^2c^2=a^2b^2
∵cosc=(a^2+b^2-c^2)/(2ab)
∴(cosc)^2=(a^4+b^4+c^4+2a^2b^2-2a^2c^2-2b^2c^2)/(4a^2b^2)
=(a^2b^2)/(4a^2b^2)
=1/4
∴cosc=±1/2.
所以∠C等于60度或120度
c²=a²+b²﹢ab=c²=a²+b²-2abcosC
cosC=-1/2
C=120°
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