Question

有大佬帮忙解一下这道题吗谢谢

Answer 1

题目都看不全，不可们就这么几个，等式根本就不成立

Answer 2

由余弦定理，
cosC=(a^2+b^2-c^2)/2ab
故(a^2+b^2)/ab=a/b+b/a=4cosC=2(a^2+b^2-c^2)/ab
解得
2c^2=a^2+b^2
另一方面，
1/6+cosC=cos(A-B)+cosC=cos(A-B)-cos(A+B)=2sinAsinB
=2sinAsinB(sinC)^2/(sinC)^2
=2ab(sinC)^2/c^2=4(sinC)^2/(2c^2/ab)
=4(sinC)^2/((a^2+b^2)/ab)
=4(sinC)^2/4cosC
=(sinC)^2/cosC
=(1-(cosC)^2)/cosC
接下来两边同乘cosC，就化成了关于cosC的一元二次方程，自己求解一下即可