2.

a, x-13=-46

=>x=(-46)+13

=>x=33

b, 4x-6=22

=>4x=22+6

=>4x=28

=>x=28:4

=>x=7

3.

a, 32=2⁵

48=2⁴.3

=>ƯCLN(32,48)=2⁴=16

16=2⁴

72=2³.3²

=>ƯCLN(16,72)=2³=8

b,

24=2³.3

60=2².3.5

=>BCNN(24,60)=2³.3.5=120

72=2³.3²

180=2².3².5

=>BCNN(72,180)=2³.3².5=360

a) x - 13  =  -   46 

    x         =  - ( 46 + 13 )                                

    x         =  - 59   

Chắc vậy

a, 36:(x–5) =  2 2

(x–5) = 9

x = 14

b, [3.(70–x)+5]:2 = 46

[3.(70–x)+5] = 92

70–x = 29

x = 41

c, 450:[41–(2x–5)] =  3 2 .5

41–(2x–5) = 10

2x–5 = 31

2x = 36

x = 18

d, 230+[ 2 4 +(x–5)] = 315. 2018 0

16+(x–5) = 315–230

x–5 = 85–16

x = 69+5

x = 74

e,  2 x + 2 x + 1  = 48

2 x .(2+1) = 48

2 x = 16 =  2 4

x = 4

f,  3 x + 2 + 3 x  = 2430

3 x . 3 2 + 1 = 2430

3 x = 2430:10 = 243 =  3 5

x = 5

mk cũng lớp 5 nhưng chưa lm cái này tuy là hs giỏi

1,53  x 72 + 15,3 x 5,2 + 1,53 x 22 - 1,53 x 46

= 110,16 + 79,56 + 33,66 - 70,38

= 153

hok tốt !

1 , 53 x 72 + 15 , 3 x 5 , 2 + 1 , 53  x 22 - 1 , 53 x 46 

= 1 , 53 x 72 + 1 , 53 x 10 x 5 , 2 + 1 , 53 x 22 - 1 , 53 x 46 

= 1 , 53 x 72 + 1 , 53 x 52 + 1 , 53 x 22 - 1 , 53 x 46 

= 1 , 53 x ( 72 + 52 + 22 - 46 ) 

= 1 , 53 x (  146 - 46 ) 

= 1 , 53 x 100 

= 153 

Chúc bạn học tốt !!! 

\(\frac{34}{11}\cdot\frac{27}{46}\cdot\frac{23}{17}\cdot\frac{22}{9}\)

\(=\frac{34\cdot27\cdot23\cdot22}{11\cdot46\cdot17\cdot9}\)

\(=\frac{2\cdot17\cdot3\cdot3\cdot3\cdot23\cdot2\cdot11}{11\cdot2\cdot23\cdot17\cdot3\cdot3}\)

\(=\frac{3\cdot2}{1}\)

\(=6\)

=))

\(\frac{34}{11}x\frac{27}{46}x\frac{23}{17}x\frac{22}{9}\)

= 2/1 x 3/2 x 1/1 x 2/1

= 6

a. x - 108 = -14

x = 94

b. 84 + x = -80

x = -164

cứu mình điiiiiiiiiiiiiiiiiiiiiiiiiiiiii mọi người!

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)

\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)

\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)

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