Nguyễn Huy TúHoàng Thị Ngọc AnhAkai Harumangonhuminhhelp me!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{a}{a+2b}=\frac{b}{b+2c}=\frac{c}{c+2a}=\frac{a+b+c}{a+2b+2+2c+1+2a}\\ =\frac{a+b+c}{\left(a+2a\right)+\left(b+2b\right)+\left(c+2c\right)}\\ =\frac{a+b+c}{3a+3b+3c}\\ =\frac{a+b+c}{3\left(a+b+c\right)}\)
Ta có:
\(a+b+c⋮a+b+c\\ \Rightarrow a+b+c⋮3\)
Vậy \(a+b+c⋮3\)
\(\dfrac{a}{a+2b}\) = \(\dfrac{b}{b+2c}\) =\(\dfrac{c}{c+2a}\)
=\(\dfrac{a+b+c}{a+2b+b+2c+c+2a}\)
=\(\dfrac{a+b+c}{3\left(a+b+c\right)}\)
= \(\dfrac{1}{3}\)
vậy a+b+c chia hết cho 3
