Bài 1 :
$9-25 = (7-3x)-(25+7)$
$⇔ 9 -25 = 7-3x-25-7$
$⇔9 =-3x$
$⇔x=-3$
Vậy $x=-3$
b) $(2x-18).(3x+12)=0$
$⇔2.(x-9).3.(x+4)=0$
$⇔(x-9).(x+4)=0$
$⇔ \left[ \begin{array}{l}x-9=0\\x+4=0\end{array} \right.$
$⇔ \left[ \begin{array}{l}x=9\\x=-4\end{array} \right.$
Vậy $x ∈\{9,-4\}$
Bài 2 :
a) $S = 1-2-3+4+5-6-7+....+2001-2002-2003+2004+2005$
$ = (1-2-3+4)+(5-6-7+8)+.....+(2001-2002-2003+2004)+2005$
$ = 0+0+....+0+2005$
$ = 2005$
Vậy $S =2005$
b) Để $5x+14 $ là bội của $2x+1$
$⇔2.(5x+4) \vdots 2x+1$
$⇔10x + 8 \vdots 2x+1$
$⇔5.(2x+1) + 3 \vdots 2x+1$
$⇔ 3 \vdots 2x+1$
$⇔2x+1 ∈ Ư(3)$
$⇔2x+1 ∈ \{-1,1,-3,3\}$
$⇔2x ∈ \{-2,0,-4,2\}$
$⇔x ∈ \{-1,0,-2,1\}$
mà : $x ∈ Z^+$
$⇒ x ∈ \{0,2\}$