Đáp án:
$\\$
`D=|x-1| +|y+2|+3`
Với mọi `x,y` ta có : \(\left\{ \begin{array}{l}|x-1|\geqslant 0 \\|y+2|\geqslant 0 \end{array} \right.\)
$↔ |x-1| + |y+2| \geqslant 0 ∀x,y$
$↔ |x-1| + |y+2| + 3 \geqslant 3 ∀ x,y$
$↔ D \geqslant 3 ∀ x,y$
`↔ min D =3`
Dấu "`=`" xảy ra khi :
`↔` \(\left\{ \begin{array}{l}|x-1|=0\\|y+2|=0\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x-1=0\\y+2=0\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=0+1\\y=0-2\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=1\\y=-2\end{array} \right.\)
Vậy `min D=3 ↔ x=1y=-2`