Đáp án: $ (x,y,z)\in\{(5,3,2), (10, 6,4)\}$
Giải thích các bước giải:
Ta có:
$\dfrac{3z-2y}{2015}=\dfrac{2x-5z}{2016}=\dfrac{5y-3x}{2017}$
$\to \dfrac{5(3z-2y)}{5\cdot 2015}=\dfrac{3(2x-5z)}{3\cdot 2016}=\dfrac{2(5y-3x)}{2\cdot 2017}$
$\to\dfrac{15z-10y}{5\cdot 2015}=\dfrac{6x-15z}{3\cdot 2016}=\dfrac{10y-6x}{2\cdot 2017}$
$\to \dfrac{15z-10y}{5\cdot 2015}=\dfrac{6x-15z}{3\cdot 2016}=\dfrac{10y-6x}{2\cdot 2017}=\dfrac{15z-10y+6x-15z+10y-6x}{5\cdot 2015+3\cdot 2016+2\cdot 2017}$
$\to \dfrac{15z-10y}{5\cdot 2015}=\dfrac{6x-15z}{3\cdot 2016}=\dfrac{10y-6x}{2\cdot 2017}=\dfrac{0}{5\cdot 2015+3\cdot 2016+2\cdot 2017}$
$\to \dfrac{15z-10y}{5\cdot 2015}=\dfrac{6x-15z}{3\cdot 2016}=\dfrac{10y-6x}{2\cdot 2017}=0$
$\to 15z-10y=6x-15z=10y-6x=0$
$\to 6x=10y=15z$
$\to \dfrac{6x}{60}=\dfrac{10y}{60}=\dfrac{15z}{60}$
$\to \dfrac{x}{10}=\dfrac{y}{6}=\dfrac{z}{4}$
$\to y=\dfrac35x, z=\dfrac25x$
Mà $yz<3x$
$\to \dfrac35x\cdot \dfrac25x<3x$
$\to \dfrac{6x^2}{25}<3x$
$\to \dfrac{6x^2}{25}-3x<0$
$\to \dfrac{2x^2}{25}-x<0$
$\to x(\dfrac{2x}{25}-1)<0$
$\to x(2x-25)<0$
$\to 0<x<\dfrac{25}{2}$
Mà $x,y,z\in Z$
$\to \dfrac35x, \dfrac25x\in Z$
$\to x\quad\vdots\quad 5$
$\to x\in\{5,10\}$
$\to (x,y,z)\in\{(5,3,2), (10, 6,4)\}$
