Ta có:A= a/b+c = b/a+c = c/a+b
=>A+1 = (a/b+c)+1 = (b/a+c)+1 = (c/a+b)+1
A+1= a+b+c/b+c = a+b+c/a+c = a+b+c/a+b
A+1= (a+b+c+a+b+c+a+b+c)/(b+c+a+c+a+b)
A+1= 3(a+b+c)/2(a+b+c)
A+1=3/2
=>A=(3/2)-1
A=1/2
Chắc thế
\(A=\frac{a+b+c}{b+ c+a+c+a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
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