Bài 1:
a, $8.25.(-125).4.12.(-2005)^0:(-3)$
$=8.(-125).25.4.12.1:(-3)$
$=-1000.100.12:(-3)$
$=-1 200 000:(-3)$
$=400 000$
b, $-128.[(-25)+89]+128.(89-125)$
$=128.25-128.89+128.89-128.125$
$0$
Bài 2:
a, $2x+28=35-(-13)$
$⇒2x+28=48$
$⇒2x=20$
$⇒x=10$
b, $2(x+1)^2=-7+15$
$⇒2(x+1)^2=8$
$⇒(x+1)^2=4$
$⇒$\(\left[ \begin{array}{l}x+1=2\\x+1=-2\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=1\\x=-3\end{array} \right.\)
Bài 3:
$\left \{ {{x^2+2>0} \atop {x-6<0}} \right.$ ⇒$\left \{ {{x^2>-2} \atop {x<6}} \right.$ ⇒$-2<x^2<36$
Vậy x∈{±1;±2;±3;±4;±5}
