a, $|x+4|=5-(-1)$
$⇔|x+4|=6$
$⇔\left[ \begin{array}{l}x+4=6\\x+4=-6\end{array} \right.$
$⇔\left[ \begin{array}{l}x=2\\x=-10\end{array} \right.$
$Vậy...$
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b, $|x-5|:2=10$
$⇔|x-5|=20$
$⇔\left[ \begin{array}{l}x-5=20\\x-5=-20\end{array} \right.$
$⇔\left[ \begin{array}{l}x=25\\x=-15\end{array} \right.$
$Vậy...$
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c, $|x-1|=-9-3$
$⇔|x-1|=-12$
Vì $|x|≤0∀x∈Z$ nên $x∈∅$
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d, $|x+2|=12+(-3)+|4|$
$⇔|x+2|=13$
$⇔\left[ \begin{array}{l}x+2=13\\x+2=-13\end{array} \right.$
$⇔\left[ \begin{array}{l}x=11\\x=-15\end{array} \right.$
$Vậy...$
