设△ABC的内角ABC的内角对边分别为a,b,c,(a+b+c)(a-b+c)=ac1.求B2.若sinAsinC=(根
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(a+b+c)(a-b+c)=3ac(a+c)²-b²=3aca²+c²-b²=ac则:cosB=(a²+c²-b²)/(2ac)=1/2得:B=60° (1)2sinAcosC-sin(A-C)=2sinAcosC-(sinAcosC-cosAsinC)=sinAcosC+cosAsinC=sin(A+C)=sinB=√3/2 (2)b=3b²=a²+c²-2accosB=a²+c²-ac=(a+c)²-3ac因为:(a+c)≥2√(ac)则:ac≤(1/4)(a+c)²得:3ac≤(3/4)(a+c)²则:(a+c)²-b²≤(3/4)(a+c)(1/4)(a+c)²≤b²a+c≤2b则:周长是a+b+c≤3b=9即三角形周长最大是9