cosC=[a^2+b^2-(a^2+b^2+ab)]/2ab=-1/2 => C=120°
2,變形得 a=b+4,a=c+8,所以最大邊為a
cosA=(b^2+c^2-a^2)/2bc=-1/2;
帶入解方程得a=14
3,由C=3B易得0 (根號2)/2
所以04,cosC=(a^2+b^2-c^)/2ab=(根號3)/2
=>a^2+b^2-c^2=(根號3)ab
=>(a+b)^2-c^2=[2+(根號3)]ab
=>(a+b)^2-c^2<=[2+(根號3)][(a+b)/2]^2
=>(a+b)^2<=4c^2/[2-(根號3)]
(先分母有理化再對2+(根號3)開根號為(根號2+根號6)/2)
=>a+b<=c*(根號6+根號2)=8+4(根號3);
(我懷疑你的c應(yīng)該為(根號6-根號2)這樣計算的最大值剛好為4)