a,|x+25|+|-y+5|=$0$ ($1$)
Nxét: |x+25| ≥ $0$ ∀ x
|-y+5| ≥ $0$ ∀ y
Từ ($1$)⇒\(\left[ \begin{array}{l}x+25=0\\-y+5=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-25\\y=5\end{array} \right.\)
Vậy ($x;y$)=($-25;5$)
b,|x-40|+|x-y+10 |≤ $0$ ($2$)
Nxét: |x-40| ≥ $0$ ∀ x
|x-y+10| ≥ $0$ ∀ y
Từ ($2$)⇒\(\left[ \begin{array}{l}x-40=0\\x-y+10=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=40\\40-y=-10⇔y=50\end{array} \right.\)
Vậy ($x;y$)=($40;50$)
