Bài 13: 

a: =>20-x=15-8+13=20

hay x=0

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.

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

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

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

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

\(a,2x-5+17=6\\ \Rightarrow2x=-6\\ \Rightarrow x=-3\\ b,\Leftrightarrow10-8+6x=-4\\ \Leftrightarrow6x=-6\Leftrightarrow x=-1\\ d,\Rightarrow-2x=-10\\ \Rightarrow x=5\)

câu c giống câu b nhó

Bài 1: 

a: BCNN(10;12)=60

b: BCNN(24;10)=120

c: BCNN(4;14;26)=364

d: BCNN(6;8;10)=120

a,(-12).x= 60-12

(-12).x=48

x=48:(-12)

x = -4

b,(-5).x + 5 = -24+6

(-5).x + 5 = 30

(-5).x      = 30-5

(-5).x      = 25

x = 25 : (-5)

x = -5

c,

$3x+36=-7x-64$

$10x=-100$

$x=-10$

$b,-5x-178=14x+145$

$-19x=323$

$x=-17$

a: =>x-3=9

=>x=12

b: =>10-x=-26

=>x=36

c: =>x:4-1=2

=>x:4=3

=>x=12

d: =>x^2=4

=>x=2 hoặc x=-2

e: =>(x-2)^2=100

=>x-2=10 hoặc x-2=-10

=>x=12 hoặc x=-8

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

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

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

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

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

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

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

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

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

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

\(a,\dfrac{6}{5}=\dfrac{18}{x}\\ \Rightarrow x=18:\dfrac{6}{5}\\ \Rightarrow x=15\\ b,\dfrac{3}{4}=\dfrac{-21}{x}\\ \Rightarrow x=-21:\dfrac{3}{4}\\ \Rightarrow x=-28\\ c,\dfrac{2}{-7}=\dfrac{18}{x}\\ \Rightarrow x=18:\dfrac{2}{-7}\\ \Rightarrow x=-63\\ d,\dfrac{-5}{2}=\dfrac{10}{-x}\\ \Rightarrow x=-10:\dfrac{-5}{2}\\ \Rightarrow x=4\)

\(a,\dfrac{6}{5}=\dfrac{18}{x}\Rightarrow6.x=5.18=90\\ \Rightarrow6.x=90\\ \Rightarrow x=15\\ b,\dfrac{3}{4}=\dfrac{-21}{x}\Rightarrow3.x=4.21=84\\ \Rightarrow x=28\)

\(a,\dfrac{6}{5}=\dfrac{18}{x}=>x=\dfrac{18.5}{6}=\dfrac{90}{6}=15\)

\(b,\dfrac{3}{4}=-\dfrac{21}{x}=>x=-\dfrac{21.4}{3}=-\dfrac{84}{3}=24\)

\(c,\dfrac{2}{-7}=\dfrac{18}{x}=>x=\dfrac{18.-7}{2}=-\dfrac{126}{2}=-63\)

\(d,-\dfrac{5}{2}=\dfrac{10}{-x}=>-x=\dfrac{10.2}{-5}=\dfrac{20}{-5}=-4=>x=4\)

a)

\(\begin{array}{l}x + \left( { - \frac{1}{5}} \right) = \frac{{ - 4}}{{15}}\\x = \frac{{ - 4}}{{15}} + \frac{1}{5}\\x = \frac{{ - 4}}{{15}} + \frac{3}{{15}}\\x = \frac{{ - 1}}{{15}}\end{array}\)                 

Vậy \(x = \frac{{ - 1}}{{15}}\).

b)

\(\begin{array}{l}3,7 - x = \frac{7}{{10}}\\x = 3,7 - \frac{7}{{10}}\\x = \frac{{37}}{{10}} - \frac{7}{{10}}\\x=\frac{30}{10}\\x = 3\end{array}\)

Vậy \(x = 3\).

c)

\(\begin{array}{l}x.\frac{3}{2} = 2,4\\x.\frac{3}{2} = \frac{{12}}{5}\\x = \frac{{12}}{5}:\frac{3}{2}\\x = \frac{{12}}{5}.\frac{2}{3}\\x = \frac{8}{5}\end{array}\)

Vậy \(x = \frac{8}{5}\)       

d)

\(\begin{array}{l}3,2:x =  - \frac{6}{{11}}\\\frac{{16}}{5}:x =  - \frac{6}{{11}}\\x = \frac{{16}}{5}:\left( { - \frac{6}{{11}}} \right)\\x = \frac{{16}}{5}.\frac{{ - 11}}{6}\\x = \frac{{ - 88}}{{15}}\end{array}\)

Vậy \(x = \frac{{ - 88}}{{15}}\).

\(a,\left(2x-5\right)+17=6\\ \Rightarrow2x-5=-11\\ \Rightarrow2x=-6\\ \Rightarrow x=-3\\ b,10-2\left(4-3x\right)=-4\\ \Rightarrow2\left(4-3x\right)=14\\ \Rightarrow4-3x=7\\ \Rightarrow3x=-3\\ \Rightarrow x=-1\\ c,24:\left(3x-2\right)=-3\\ \Rightarrow3x-2=-8\\ \Rightarrow3x=-6\\ \Rightarrow x=-2\\ d,5-2x=-17+12\\ \Rightarrow5-2x=-5\\ \Rightarrow2x=10\\ \Rightarrow x=5\)

a: =>2x-5=-11

=>2x=-6

hay x=-3

b: =>2(4-3x)=14

=>4-3x=7

=>3x=-3

hay x=-1

c: =>3x-2=-8

=>3x=-6

hay x=-2

a) 2x-5+17=6

2x+12=6

2x=12-6

2x=6

x=6:2

x-3

b) 10-8+6x=-4

2+6x=-4

6x=-4-2

6x=-6

x=(-6):6

x=-1

c) 3x-2=24:(-3)

3x-2=-8

3x=-8+2

3x=-6

x=-6:3

x=-2

d) 5-2x=-5

-2x=-5-5

-2x=-10

x=-10:(-2)

x=5