三角形ABC中,2（a^2+b^2-c^2)=3ab,若c=2,求ABC面积的最大值

2(a^2+b^2-c^2)=3ab
a^2+b^2-c^2=3ab/2
根据余弦定理：cosC=(a^2+b^2-c^2)/(2ab)=(3ab/2)/(2ab)=3/4
sinC=根号(1-cos^2C)=根号7/4
S△ABC=1/2absinC=1/2ab*根号7/4 = (ab根号7) /8
将c=2代入2(a^2+b^2-c^2)=3ab
2(a^2+b^2-4)=3ab
2(a^2+b^2)-8=3ab
2(a^2-2ab+b^2)+4ab-8=3ab
ab=8-2(a-b)^2
当a=b时,ab取最大值8
∴S△ABCmax= (ab根号7) /8 = 8*根号7 /8 = 根号7