Question

[Hânn Ngọc:))]

làm hộ e nhé!!:>

Answer 1

[Nguyễn Việt Lâm]

1.

Nếu \(a+b+c=0\Rightarrow B=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{\left(-c\right)\left(-a\right)\left(-b\right)}{abc}=-1\)

Theo tính chất dãy tỉ số bằng nhau:

\(\dfrac{a+b-2017c}{c}=\dfrac{b+c-2017a}{a}=\dfrac{c+a-2017b}{b}=\dfrac{a+b-2017c+b+c-2017a+c+a-2017b}{c+a+b}\)

\(=\dfrac{-2015\left(a+b+c\right)}{a+b+c}=-2015\)

\(\Rightarrow\left\{{}\begin{matrix}a+b-2017c=-2015c\\a+c-2017b=-2015b\\b+c-2017a=-2015a\end{matrix}\right.\)

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

\(\Rightarrow\left\{{}\begin{matrix}2c-b=2b-c\\b+c=2a\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}3b=3c\\b+c=2a\end{matrix}\right.\)

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

\(\Rightarrow a=b=c\)

Do đó:

\(B=\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)=2.2.2=8\)

Answer 2

[Trên con đường thành công không có....]