14′

(2)由

得  accosB=b²-(a-c)²                                           9′

即accosB=a²+c²-2accosB-(a²+c²-2ac)                             11′

∴最大角为C,有                 5′

C=120°                                                      7′

18． 在中有sinA:sinB=3:5  

∴a : b=3 : 5                                                     2′

令a = 3k  (k>0)    ∴ b = 5k

∵a,b,c成等差数列    ∴ c =7k

综上所述：                                            

17．问题转化为对任意实数恒成立．