Lúc nãy, cô còn dạy học nên giờ cô mới giảng cho em được nhé.

B = (1 - \(\dfrac{1}{2}\))\(\times\)(1 - \(\dfrac{1}{3}\))\(\times\)(1 - \(\dfrac{1}{4}\))\(\times\)(1-\(\dfrac{1}{5}\))\(\times\)...\(\times\)(1- \(\dfrac{1}{2003}\))\(\times\)(1-\(\dfrac{1}{2004}\))

B = \(\dfrac{2-1}{2}\)\(\times\)\(\dfrac{3-1}{3}\)\(\times\)\(\dfrac{4-1}{4}\)\(\times\)\(\dfrac{5-1}{5}\)\(\times\)...\(\times\)(\(\dfrac{2003-1}{2003}\))\(\times\)(\(\dfrac{2004-1}{2004}\))

B = \(\dfrac{1}{2}\)\(\times\)\(\dfrac{2}{3}\)\(\times\)\(\dfrac{3}{4}\)\(\times\)\(\dfrac{4}{5}\)\(\times\)...\(\times\)\(\dfrac{2002}{2003}\)\(\times\)\(\dfrac{2003}{2004}\)

B = \(\dfrac{2\times3\times4\times...\times2003}{2\times3\times4\times...\times2003}\)\(\times\) \(\dfrac{1}{2004}\)

B = \(\dfrac{1}{2004}\)

\(\dfrac{2}{5}+\dfrac{3}{5}:\left(\dfrac{3}{5}+-\dfrac{2}{3}\right)-\dfrac{5}{2}\)

\(=\dfrac{2}{5}+\dfrac{3}{5}:\left(-\dfrac{1}{15}\right)-\dfrac{5}{2}\)

\(=\dfrac{2}{5}+\left(-9\right)-\dfrac{5}{2}\)

\(=-\dfrac{21}{10}+-9\)

\(=-\dfrac{111}{10}\)

\(\dfrac{2}{5}+\dfrac{3}{5}\):(\(\dfrac{3}{5}\) - \(\dfrac{2}{3}\)) - \(\dfrac{5}{2}\)

= \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\):(-\(\dfrac{1}{15}\)) - \(\dfrac{5}{2}\)

= \(\dfrac{2}{5}\) - 9 - \(\dfrac{5}{2}\)

= \(\dfrac{4}{10}\) - \(\dfrac{90}{10}\) - \(\dfrac{25}{10}\) 

= \(-\dfrac{111}{10}\)

`(1/8 +(-3)/4) : 5/6 +1/2`

`= (1/8 + (-6)/8) : 5/6 +1/2`

`= -5/8 : 5/6 +1/2`

`= -5/8 xx 6/5 +1/2`

`= -30/40 +1/2`

`= -3/4+ 2/4`

`= -1/4`

\(\left(\dfrac{1}{8}+\dfrac{-3}{4}\right):\dfrac{5}{6}+\dfrac{1}{2}\)

\(=\left(\dfrac{1}{8}+\dfrac{-6}{8}\right)\times\dfrac{6}{5}+\dfrac{1}{2}\)

\(=\dfrac{-5}{8}\times\dfrac{6}{5}+\dfrac{1}{2}\)

\(=\dfrac{-3}{4}+\dfrac{1}{2}\)

\(=\dfrac{-3}{4}+\dfrac{2}{4}\)

\(=\dfrac{-1}{4}\)

Chúc bạn học tốt

\(83\times14+27\times17+83\times13\)

\(=83\times\left(14+13\right)+27\times17\)

\(=83\times27+27\times17\)

\(=\left(83+17\right)\times27\)

\(=100\times27\)

\(=2700\)

Chúc bạn học tốt

$83\times 14+27\times 17+83\times 13$

$=83\times \left(14+13\right)+27\times 17$

$=83\times 27+27\times 17$

$=\left(83+17\right)\times 27$

$=100\times 27$

$=2700$

Chúc bạn học tốt

F = 7 + 72 + 73 + 74 + ..... + 7100 

F= 7+(1+7)+73+(1+7)+...+799+(1+7)

F = 7x8+73x8+...+799x8

F= 8x(7+73+...+799)

mà 8 chia hết 8 => 8(7+73+...+799) chia hết 8

Vậy F chia hết cho 8

2)

\(F=7+7^2+7^3+7^4+...+7^{100}\\ F=7\cdot\left(1+7\right)+7^3\cdot\left(1+7\right)+.....+7^{99}\cdot\left(1+7\right)\\F=7\cdot8+7^3\cdot8+.....+7^{99}\cdot8\\ F=8\cdot\left(7+7^3+....+7^{99}\right)\\ =>F⋮8\) 

a) 252 - 84 : 21 + 7

= 252 - 4 + 7

= 248 + 7 

= 255

b) ( 252 - 84 ) : 21 + 7

= 168 : 21 + 7

= 8 + 7 = 15

252 - 84 : ( 21 + 7 )

= 252 - 84 : 28

= 252 - 3

= 249

a) 252 - 84 : 21 + 7

= 252 - 4 + 7

= 248 + 7

= 255

b) (252 - 84) : 21 + 7

= 168 : 21 + 7

= 8 + 7

= 15

( 252 - 84) : (21 + 7)

= 168 : 28

= 6

 252 - (84 : 21 + 7)

= 252 - 11

= 241

 252 - 84 : (21 + 7)

= 252 - 84 : 28

= 252 - 3

= 249

a) Ta có A = 1 + 2¹ + 2² + ... + 2²⁰²¹

           2A = 2¹ + 2² + 2³ + ... + 2²⁰²²

Vậy 2A = 2¹ + 2² + 2³ + ... + 2²⁰²²

b) 2A - A = ( 2¹ + 2² + 2³ + ... + 2²⁰²² ) - ( 1 + 2¹ + 2² + ... + 2²⁰²¹ )

           A = 2²⁰²² - 1

Vậy A = 2²⁰²² - 1

a)

\(A=1+2^1+2^2+2^3+...+2^{2020}+2^{2021}\)

\(2A=2^1+2^2+2^3+2^4+...+2^{2021}+2^{2022}\)

b)

\(2A=2^1+2^2+2^3+...+2^{2022}\)

\(2A-A=\left(2^1+2^2+2^3+...+2^{2022}\right)-\left(1+2^1+2^2+....+2^{2021}\right)\)

\(A=2^{2022}-1\)

=> đpcm

\(A=1+2+2^2+2^3+...+2^{2020}+2^{2021}\)

\(\Rightarrow2A=2+2^2+2^3+2^4+...+2^{2021}+2^{2022}\)

\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{2021}+2^{2022}\right)-\left(1+2+2^2+2^3+...+2^{2020}+2^{2021}\right)\)

\(\Rightarrow A=\left(2-2\right)+\left(2^2-2^2\right)+...+\left(2^{2021}-2^{2021}\right)+\left(2^{2022}-1\right)\)

\(\Rightarrow A=2^{2022}-1\)

\(12\times\left(x-4\right)=0\)

\(\Rightarrow x-4=0:12\)

\(\Rightarrow x-4=0\)

\(\Rightarrow x=0+4\)

\(\Rightarrow x=4\)

12 x (X - 4 )= 0

         X - 4 = 0: 12

          X- 4 = 0

               X = 0 + 4 

               X = 4.

Vậy : X = 4 .

\(12\left(x-4\right)=0\)

\(\Rightarrow x-4=0\)

\(\Rightarrow x=4\)