ĐKXĐ : `x \ne 0`
\(\left\{ \begin{array}{l}\dfrac{x+1}x+|x-y|=3\\\dfrac1x-2|x-y|=-1\end{array} \right.\)
`⇔`\(\left\{ \begin{array}{l}\dfrac{2(x+1)}x+2|x-y|=6(1)\\\dfrac1x-2|x-y|=-1(2)\end{array} \right.\)
`(1) + (2)` được :
`⇔ (2(x+1))/x + 2|x-y| + 1/x - 2|x-y| = 6-1`
`⇔ (2x+2)/x + 1/x = 5`
`⇔ (2x+3)/x = 5`
`⇔ 2x + 3 = 5x`
`⇔ 2x - 5x = -3`
`⇔ -3x = -3`
`⇔ x = 1(TM)`
Thay `x = 1` vào pt `(2)` , ta được :
`⇔ 1/1 - 2|1-y| = -1`
`⇔ 1 - 2|1-y| = -1`
`⇔ 2|1-y| = 2`
`⇔ |1-y| = 1`
`⇔`\(\left[ \begin{array}{l}1-y=1\\1-y=-1\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}y=0\\y=-2\end{array} \right.\)
Vậy `x = 1` và `y = 0` hoặc `y = -2`
