`|x - 2| + |x - 5| + |x - y + 1| + |x - 2017| = 2015`
`<=> |x - 2| + |x - 5| + |x - y + 1| + |2017 - x| = 2015`
`<=> (|x - 2| + |2017 - x|) + (|x - 5| + |x - y + 1|) = 2015`
`+)` Áp dụng bất đẳng thức giá trị tuyệt đối `|x| + |y| >= |x + y|`, ta có:
`|x - 2| + |2017 - x| >= |x - 2 + 2017 - x| = |2015| = 2015`
`=> |x - 2| + |2017 - x| >= 2015(**)`
`+)` Ta có:
`|x - 5| >= 0 AA x, |x - y + 1| >= 0 AA x, y`
`=> |x - 5| + |x - y + 1| >= 0 AA x, y(***)`
Từ `(**)` và`(***)`
` => (|x - 2| + |2017 - x|) + (|x - 5| + |x - y + 1|) >= 2015`
Dấu "`=`" xảy ra
`<=> |x - 5| + |x - y + 1| = 0`
`<=>` \(\left\{\begin{matrix}|x - 5| = 0\\|x - y + 1| = 0\end{matrix}\right.\)
`<=>` \(\left\{\begin{matrix}x - 5 = 0\\x - y + 1 = 0\end{matrix}\right.\)
`<=>` \(\left\{\begin{matrix}x = 5\\x - y = -1\end{matrix}\right.\)
`<=>` \(\left\{\begin{matrix}x = 5\\5 - y = -1\end{matrix}\right.\)
`<=>` \(\left\{\begin{matrix}x = 5\\y = 6\end{matrix}\right.\)
Vậy `x = 5, y = 6`
