作业帮(sinB)^2+(sinC)^2-(sinA)^2=根号3sinAsinC求B
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/29 21:59:36
(sinB)^2+(sinC)^2-(sinA)^2=根号3sinAsinC求B
(sinB)^2+(sinC)^2-(sinA)^2=根号3sinAsinC
求B
(sinB)^2+(sinC)^2-(sinA)^2=根号3sinAsinC求B
错了吧
应该是(sinA)^2+(sinC)^2-(sinB)^2=根号3sinAsinC
a/sinA=b/sinB=c/sinC=k
sinA=a/k,sinB=b/k,sinC=c/k
代入,把k约分
a^2+c^2-b^2=√3ac
cosB=(a^2+c^2-b^2)/2ac=√3/2
B=30度
1/(a-b)+1/(b-c)-4/(a-c)
=(b-c+a-b)/(a-b)(b-c)-4/(a-c)
=(a-c)/(a-b)(b-c)-4/(a-c)
(a-c)^2-4(a-b)(b-c)
=a^2-2ac+c^2-4ab+4ac+4b^2-4bc
=a^2+2ac+c^2-4ab+4b^2-4bc
=(a+c)^2-4b(a+c)+4b^...
全部展开
1/(a-b)+1/(b-c)-4/(a-c)
=(b-c+a-b)/(a-b)(b-c)-4/(a-c)
=(a-c)/(a-b)(b-c)-4/(a-c)
(a-c)^2-4(a-b)(b-c)
=a^2-2ac+c^2-4ab+4ac+4b^2-4bc
=a^2+2ac+c^2-4ab+4b^2-4bc
=(a+c)^2-4b(a+c)+4b^2
=(a+c-2b)^2>=0
所以(a-c)^2〉=4(a-b)(b-c)
a>b>c
所以a-c>0,a-b>0,b-c>0
所以两边除(a-c)(a-b)(b-c)
(a-c)/(a-b)(b-c)>=4/(a-c)
所以1/(a-b)+1/(b-c)-4/(a-c)〉=0
所以1/(a-b)+1/(b-c)>=4/(a-c)
收起