1,2a^2=2b^2+bc+2c^2+bc
(正弦定理得)
2b^2+2c^2-2a^2=-2bc
b^2+c^2-a^2=-bc
则cosA=-1/2
A=120度
2f(x)非零函数f(a+b)=f(a)f(b)
且当x1
固当b1,f(a+b)=f(a)f(b)>f(a)且
1,2a^2=2b^2+bc+2c^2+bc
(正弦定理得)
2b^2+2c^2-2a^2=-2bc
b^2+c^2-a^2=-bc
则cosA=-1/2
A=120度
2f(x)非零函数f(a+b)=f(a)f(b)
且当x1
固当b1,f(a+b)=f(a)f(b)>f(a)且