Question

ai giúp mình với ạ cảm ơi

Answer 1

\(\dfrac{3a+b+c}{a}=\dfrac{a+3b+c}{b}=\dfrac{a+b+3c}{c}\)

Áp dụng t/c DTSBN, ta có:

\(=\dfrac{3a+b+c+a+3b+c+a+b+3c}{a+b+c}=\dfrac{5\left(a+b+c\right)}{a+b+c}=5\)

\(\Rightarrow\dfrac{3a+b+c}{a}-3=\dfrac{a+3b+c}{b}-3=\dfrac{a+b+3c}{c}-3\)

\(=\dfrac{b+c}{a}=\dfrac{a+c}{b}=\dfrac{a+b}{c}=2\\ \Rightarrow\left\{{}\begin{matrix}2a=b+c\\2b=a+c\\2c=a+b\end{matrix}\right.\Leftrightarrow a=b=c\)

\(N=\dfrac{\left(a+a+a\right)^3}{a.a.a}=\dfrac{\left(3a\right)^3}{a^3}=\dfrac{27a^3}{a^3}=27\)