ta có:
\(\left|x-1\right|+\left|x-2\right|+\left|y-3\right|+\left|x-4\right|\)
\(=\left|x-1\right|+\left|x-2\right|+\left|y-3\right|+\left|4-x\right|\)
\(\ge\left|x-1+4-x\right|+\left|x-2\right|+\left|y-3\right|\)
\(=3+\left|x-2\right|+\left|y-3\right|\)
\(\ge3\)
Dấu "=" xả ra khi \(\hept{\begin{cases}\left(x-1\right)\left(4-x\right)\ge0\\\left|x-2\right|=0\\\left|y-3\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}1\le x\le4\cdot\\x=2\left(TM\cdot\right)\\y=3\end{cases}}\)
Vậy \(x=2;y=3\)
(x-1) + (x-2) + (x-3) + (x-4) = 3
(x+x+x+x) - (1+2+3+4) = 3
X x 4 - 10 = 3
X x 4 = 3 + 10
X x 4 = 13
x = 13 : 4
x = \(\frac{13}{4}\)
