a, 60

b, 15

c, 11

d, 11

e, 110

g, 5

a (x-12)+48=96

x-12=96-48=48

x=12+48=60

b56-(17+x)=24

17+x=56-24

17+x=32

x=31-17=15

c 5x+57=112

5x=112-57=55

x=11

d 87-2x=65

2x=87-65=22

x=11

e (x-46):32=2

x-46=64

x=110

g 120-3(x+1)=102

3(x+1)=120-102=18

x+1=6

x=5

a: Ta có: \(\left(x-12\right)+48=96\)

\(\Leftrightarrow x-12=48\)

hay x=60

b: ta có: \(56-\left(x+17\right)=24\)

\(\Leftrightarrow x+17=32\)

hay x=15

c: Ta có: \(5x+57=112\)

\(\Leftrightarrow5x=55\)

hay x=11

d: Ta có: 87-2x=65

\(\Leftrightarrow2x=22\)

hay x=11

`3/7+2/5+4/7`

`=(3/7+4/7)+2/5`

`=1+2/5`

`=5/5+2/5`

`=7/5`

`20/34+4/9+7/17`

`=(20/34+7/17)+4/9`

`=(20/34+14/34)+4/9`

`=34/34+4/9`

`=1+4/9`

`=9/9+4/9`

`=13/9`

`48xx2+96+48xx6`

`=48xx2+48xx2+48xx6`

`=48xx(2+2+6)`

`=48xx10`

`=480`

câu thứ hai thì làm như này em nhé:

    \(\dfrac{20}{34}\) + \(\dfrac{4}{9}\) + \(\dfrac{7}{17}\)

= (\(\dfrac{20}{34}\) + \(\dfrac{7}{17}\)) + \(\dfrac{4}{9}\)

=  (\(\dfrac{10}{17}\) + \(\dfrac{7}{17}\)) + \(\dfrac{4}{9}\)

= 1 + \(\dfrac{4}{9}\)

= \(\dfrac{13}{9}\)

\(\dfrac{x\times48-120\times48}{72\times48-22\times48}=1\\ \dfrac{\left(x-120\right)\times48}{\left(72-22\right)\times48}=1\\ \dfrac{x-120}{50}=1\\ \dfrac{x-120}{50}=\dfrac{50}{50}\\ x-120=50\\ x=50+120\\ x=170\)

(1-2+3-4+5-6+7-8+...+101-102+103) * x-120=2012

[ -1 + ( -1 ) ... ( -1 ) + 103] * x -120 = 2012

( -1 x 51 + 103 ) * x - 120 = 2012

( -51 +103) * x = 2012 + 120

52 * x =2132

x= 2132 : 52

x= 41

 x - (28 - x) = x - 24x−(28−x)=x−24.

\(\left[3^4-\left(8^2+14\right):13\right]\cdot x=5^3+10^2\)

\(\Rightarrow\left[81-\left(64+14\right):13\right]\cdot x=125+100\)

\(\Rightarrow\left(81-78:13\right)\cdot x=225\)

\(\Rightarrow\left(81-6\right)\cdot x=225\)

\(\Rightarrow75\cdot x=225\)

\(\Rightarrow x=225:75\)

\(\Rightarrow x=3\)

Bài 4:

a: =>7/x-5=2

=>x-5=7/2

=>x=17/2

b: =>1-2x=-5

=>2x=6

=>x=3

c: =>2x-3=5 hoặc 2x-3=-5

=>2x=8 hoặc 2x=-2

=>x=-1 hoặc x=4

d: =>2(x+1)^2+17=21

=>2(x+1)^2=4

=>(x+1)^2=2

=>\(x+1=\pm\sqrt{2}\)

=>\(x=\pm\sqrt{2}-1\)