Ta có:

\(\dfrac{12x-15y}{2017}=\dfrac{20z-12x}{2018}=\dfrac{15y-20z}{2019}\)

\(=\dfrac{12x-15y+20z-12x+15y-20z}{2017+2018+2019}\)

\(=\dfrac{0}{2017+2018+2019}=0\)

\(\Rightarrow\left\{{}\begin{matrix}12x-15y=0\\20z-12x=0\\15y-20z=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}12x=15y\\20z=12x\\15y=20z\end{matrix}\right.\)\(\Rightarrow12x=15y=20z\)

\(\Rightarrow\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}\)

Áp dụng tích chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}=\dfrac{x+y+z}{5+4+3}=\dfrac{48}{12}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=4.5=20\\y=4.4=16\\z=4.3=12\end{matrix}\right.\)

Vậy ...

Lời giải:

Với \(x=y=z=-1\) thì:

\(B=(-1)(-1)^2(-1)^3+(-1)^2(-1)^3(-1)^4+(-1)^3(-1)^4(-1)^5+...+(-1)^{2017}(-1)^{2018}(-1)^{2019}\)

\(=(-1)^{1+2+3}+(-1)^{2+3+4}+(-1)^{3+4+5}+...+(-1)^{2017+2018+2019}\)

\(=(-1)^6+(-1)^{9}+(-1)^{12}+...+(-1)^{6054}\)

\(=[(-1)^6+(-1)^{12}+(-1)^{18}+...+(-1)^{6054}]+[(-1)^9+(-1)^{15}+...+(-1)^{6051}]\)

\(=\underbrace{1+1+..+1}_{1009}+\underbrace{[(-1)+(-1)+..+(-1)]}_{1008}\)

\(=1009-1008=1\)