a: Ta có: \(7x+25=144\)

\(\Leftrightarrow7x=119\)

hay x=17

b: Ta có: \(33-12x=9\)

\(\Leftrightarrow12x=24\)

hay x=2

c: Ta có: \(128-3\left(x+4\right)=23\)

\(\Leftrightarrow3\left(x+4\right)=105\)

\(\Leftrightarrow x+4=35\)

hay x=31

d: Ta có: \(71+\left(726-3x\right)\cdot5=2246\)

\(\Leftrightarrow5\left(726-3x\right)=2175\)

\(\Leftrightarrow726-3x=435\)

\(\Leftrightarrow3x=291\)

hay x=97

e: Ta có: \(720:\left[41-\left(2x+5\right)\right]=40\)

\(\Leftrightarrow41-\left(2x+5\right)=18\)

\(\Leftrightarrow2x+5=23\)

\(\Leftrightarrow2x=18\)

hay x=9

Bn cần bài nào vậy

f: Ta có: \(10-4x+120:8=16+1\)

\(\Leftrightarrow4x=17-25=-8\)

hay x=-2

g: Ta có: \(x+9x+7x+5x=2244\)

\(\Leftrightarrow22x=2244\)

hay x=102

h: Ta có: \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)

\(\Leftrightarrow100x+5050=5750\)

\(\Leftrightarrow100x=700\)

hay x=7

a: \(\Leftrightarrow14x=56\)

hay x=4

Bài 13: 

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

hay x=0

a: =>19/23>19/x>19/29

=>\(x\in\left\{24;25;26;27;28\right\}\)

b: =>88/132<88/x<88/128

=>132>x>128

=>\(x\in\left\{131;130;129\right\}\)

c: =>\(\left\{{}\begin{matrix}\dfrac{4}{x}-\dfrac{x}{8}< 0\\\dfrac{x}{8}-\dfrac{5}{x}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{32-x^2}{8x}< 0\\\dfrac{x^2-40}{8x}< 0\end{matrix}\right.\)

=>32<x^2<40

=>x=6

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

⇒\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)

⇒\(2x+1=21\)

\(2x=21-1\)

\(2x=20\)

⇒\(x=10\)

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

a) 

\(\left(x+1\right)\left(y-2\right)=5\\ \Rightarrow\left(x+1\right),\left(y-2\right)\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)

Ta có bảng:

Vậy \(\left(x;y\right)=\left(0;7\right),\left(-2;-3\right),\left(4;3\right),\left(-6;1\right)\)

b) 

\(\left(x-5\right)\left(y+4\right)=-7\\ \Rightarrow\left(x-5\right),\left(y+4\right)\inƯ\left(-7\right)=\left\{1;-1;7;-7\right\}\)

Ta có bảng:

Vậy \(\left(x;y\right)=\left(6;-11\right),\left(4;3\right),\left(12;-5\right),\left(-2;-3\right)\)

e) 

\(x-\left(17-8\right)=5+\left(10-3x\right)\\ \Rightarrow x-9=5+10-3x\\ \Rightarrow x+3x=5+10+9\\ \Rightarrow4x=24\\ \Rightarrow x=\dfrac{24}{4}=6\)

Vậy \(x=6\)