$a,|x+5|=4$
$⇒x+5=±4$
\(⇒\left[ \begin{array}{l}x+5=4⇒x=-1\\x+5=-4⇒x=-9\end{array} \right.\)
vậy $x∈${$-1;-9$}
$b,1<|x-2|<4$
$⇒|x-2|∈${$2;3$}
\(⇒\left[ \begin{array}{l}|x-2|=2⇒\left[ \begin{array}{l}x-2=2⇒x=4\\x-2=-2⇒x=0\end{array} \right.\\|x-2|=3⇒\left[ \begin{array}{l}x-2=3⇒x=5\\x-2=-3⇒x=-1\end{array} \right.\end{array} \right.\)
Vậy $x∈${$0;-1;4;5$}
$c,|x+4|<3$
vì $|x+4|≥0$
$⇒ 0≤|x+4|<3$
$⇒|x+4|∈${$0;1;2$}
$⇒\left[\begin{array}{ccc}|x+4|=0⇒x+4=0⇒x=-4\\|x+4|=1⇒\left[ \begin{array}{l}x+4=1⇒x=-3\\x+4=-1⇒x=-5\end{array} \right.\\|x+4|=2⇒\left[ \begin{array}{l}x+4=2⇒x=-2\\x+4=-2⇒x=-6\end{array} \right.\end{array}\right]$
$d,|x-14+17|+|y+10-12|≤0$
ta thấy : $|x-14+17|;|y+10-12|≥0∀x∈Z$
$⇒ |x-14+17|+|y+10-12|≥0(1)$
$TH1:|x-14+17|+|y+10-12|<0$
từ $(1),⇒(KTM)$
$TH2:|x-14+17|+|y+10-12|=0$
từ $(1),⇒\left \{ {{|x-14+17|=0⇒x-14+17=0⇒x=-3} \atop {|y+10-12|=0⇒y+10-12⇒y=2}} \right.$
Vậy $x=-3;y=2$
