g: x:(-7)-6=-15

=>x:(-7)=-15+6=-9

=>\(x=\left(-9\right)\left(-7\right)=9\cdot7=63\)

h: \(x:12+\left(-12\right)=-37\)

=>\(x:12=\left(-37\right)-\left(-12\right)=-37+12=-25\)

=>\(x=-25\cdot12=-300\)

i: \(x:\left(-25\right)+81=85\)

=>\(x:\left(-25\right)=85-81=4\)

=>\(x=4\cdot\left(-25\right)=-100\)

k: \(\left(-31\right)\cdot x=-93\)

=>\(x=\dfrac{-93}{-31}=\dfrac{93}{31}\)

=>x=3

l: 5x+1=-4

=>5x=-4-1=-5

=>\(x=-\dfrac{5}{5}=-1\)

m: \(\left(-12\right)\cdot x+34=10\)

=>\(\left(-12\right)\cdot x=10-34=-24\)

=>12x=24

=>x=24/12=2

a: =>y/15=-2/3

hay y=-10

b: 2/x=x/18

nên \(x^2=36\)

hay \(x\in\left\{6;-6\right\}\)

c: x/9=16/x

nên \(x^2=144\)

hay \(x\in\left\{12;-12\right\}\)

a)x=-10
b)x=-6

c)x=-12 nhé

\(a,\left(x+12\right)\left(x-6\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+12>0\\x-6>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+12< 0\\x-6< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-12\\x>6\end{matrix}\right.\\\left\{{}\begin{matrix}x< -12\\x< 6\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x>6\\x< -12\end{matrix}\right.\)

\(b,\left(10-x\right)\left(3-x\right)< 0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}10-x< 0\\3-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}10-x>0\\3-x< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>10\\x< 3\left(vô.lí\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x< 10\\x>3\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x< 10\\x>3\end{matrix}\right.\)

\(a,\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+12>0\\x-6>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+12< 0\\x-6< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>6\\x< -12\end{matrix}\right.\\ \Rightarrow x\in\left\{...;-15;-14;-13;7;8;9;...\right\}\\ b,\Rightarrow\left(x-10\right)\left(x-3\right)< 0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-10>0\\x-3< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-10< 0\\x-3>0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>10;x< 3\left(\text{loại}\right)\\3< x< 10\end{matrix}\right.\\ \Rightarrow x\in\left\{4;5;6;7;8;9\right\}\)

\(a,\left(x+12\right)\left(x+6\right)>0\) \(khi\) \(x>6\Rightarrow x\in\left\{7,8,9,...\right\}\)

\(b,\left(10-x\right)\left(3-x\right)< 0\) \(khi\) \(x< 10\Rightarrow x\in\left\{9,8,7,...\right\}\)

Bài 2: 

a: \(\Leftrightarrow x-7=36\)

hay x=43

a,\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)

⇒\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)

⇒\(2x+1=21\)

\(2x=21-1\)

\(2x=20\)

⇒\(x=10\)

Lời giải:
a. 

$(-2)x-(-21)=15$

$-2x+21=15$

$-2x=15-21=-6$

$x=(-6):(-2)=3$

b.

$(3x-2^2).7^3=7^4$
$3x-2^2=7^4:7^3=7$

$3x-4=7$

$3x=11$

$x=\frac{11}{3}$

a: =>-12x=60-12=48

=>x=-4

b: =>-5x=24+6-5=25

=>x=-5

Chi tiết nha

Lời giải:
a.

$(-12).x=(-15)(-4)-12$

$(-12)x=60-12=48$

$x=48:(-12)=-4$

b.

$(-5)x+5=(-3)(-8)+6$

$(-5)x+5=24+6=30$

$(-5)x=30-5=25$

$x=25:(-5)=-5$

a,-4/7=x/21
-12/21 = x/21
x= -12 

b,(x-3)/15=1/-5
x - 3 = -1/5 * 15
x - 3 = -3
x = 0

c,.(3x+8)/-12=-5/30
=> 3x + 8 = 2
=> 3x=-6
=>x=-2

b: -7<x<7