(1)ab>c²
∴ cosC=(a²+b²-c²)/(2ab)>(2ab-c²)/(2ab)>ab/(2ab)=1/2
∴ C2c
∴ cosC=(a²+b²-c²)/(2ab)
=(4a²+4b²-4c²)/(8ab)
>[4a²+4b²-(a+b)²)/(8ab)
=[3a²+3b²-2ab]/(2ab)
≥(4ab)/(8ab)
=1/2
∴ C0
∴ C是锐角
(4)举反例 a=b=c=2
满足(a+b)c=2ab,是等边三角形,C=π/3
(1)ab>c²
∴ cosC=(a²+b²-c²)/(2ab)>(2ab-c²)/(2ab)>ab/(2ab)=1/2
∴ C2c
∴ cosC=(a²+b²-c²)/(2ab)
=(4a²+4b²-4c²)/(8ab)
>[4a²+4b²-(a+b)²)/(8ab)
=[3a²+3b²-2ab]/(2ab)
≥(4ab)/(8ab)
=1/2
∴ C0
∴ C是锐角
(4)举反例 a=b=c=2
满足(a+b)c=2ab,是等边三角形,C=π/3