1﹥∵b^2+c^2=a^2+bc
∴COS∠A=(b^2+c^2-a^2)/(2bc)=0.5 ∴A=60º
∴原式=2SinBCosC-Sin(B-C)
=2SinBCosC-(SinBCosC-SinCCosB)
=SinBCosC+SinCCosB
=Sin(B+C)=SinA=Sin60º=√3/2
2>∵a=2 ∴b²+c²=a²+bc=4+bc
∴(b+c)²-4=3bc≤3(b+c)²/4
∴(b+c)²≤16 b+c≤4
∴a+b+c≤6
故 三角形ABC周长的最大值为6
1﹥∵b^2+c^2=a^2+bc
∴COS∠A=(b^2+c^2-a^2)/(2bc)=0.5 ∴A=60º
∴原式=2SinBCosC-Sin(B-C)
=2SinBCosC-(SinBCosC-SinCCosB)
=SinBCosC+SinCCosB
=Sin(B+C)=SinA=Sin60º=√3/2
2>∵a=2 ∴b²+c²=a²+bc=4+bc
∴(b+c)²-4=3bc≤3(b+c)²/4
∴(b+c)²≤16 b+c≤4
∴a+b+c≤6
故 三角形ABC周长的最大值为6