2016x(x - 2017/2016)=0

2016x^2-2017x=0

x(2016x-2017)=0

suy ra : x=0 hoặc 2016x - 2017 = 0

           x = 0 hoặc 2016x          =2017

           x = 0 hoặc x                 = 2017/2016

2) 5(x-2)+3x(2-x)=0

5x(x-2)-3x(2-x)=0

(x-2)(5--3x)=0

suy ra : x-2=0 hoặc 5-3x=0

           x=2  hoặc 3x   =5

          x=2 hoặc x =5/3

a 2016 x ( 2015 + 2017 - 4030 ) = 2016 x 2 = 4032

Có: \(\left(2x-1\right)^{2016}\ge0;\left(y-\frac{2}{5}\right)^{2016}\ge0;\left|x+y+z\right|\ge0\forall x;y;z\)

Mà theo đề bài: \(\left(2x-1\right)^{2016}+\left(y-\frac{2}{5}\right)^{2016}+\left|x+y+z\right|=0\)

\(\Rightarrow\begin{cases}\left(2x-1\right)^{2016}=0\\\left(y-\frac{2}{5}\right)^{2016}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}2x-1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}2x=1\\y=\frac{2}{5}\\x+y+z=0\end{cases}\)

\(\Rightarrow\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{-9}{10}\end{cases}\)

Vậy \(x=\frac{1}{2};y=\frac{2}{5};z=\frac{-9}{10}\)