1x2x3x...2018x2019 - 1x2x3x..2018 - 1x2x3x4x...x2017x2018² 

= 1x2x3x...x2018x(2019 - 1 - 2018)

= 1x2x3x...x2018x0

= 0

A = (-1)(-1)^2(-1)^3...(-1)^2019

A = (-1)^1+2+3+...+2019

A = (-1)^2039190

A = 1

S = 1.2.3 + 2.3.4 + 3.4.5 + ... + 2018.2019.2020

4S = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + .... + 2018.2019.2020.4

4S = 1.2.3.4 + 2.3.4.(5 - 1) + 3.4.5.(6 - 2) + ... + 2018.2019.2020.(2021 - 2017)

4S = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 2018.2019.2020.2021 - 2017.2018.2019

4S = 2018.2019.2020.2021

S = 2018.2019.2020.2021 : 4 = ...

cảm ơn bạn nhiều nhé

Giải:

a) \(75\%+1,2-2+\dfrac{1}{5}+2018^0\) 

=\(\dfrac{3}{4}+\dfrac{6}{5}-2+\dfrac{1}{5}+1\) 

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

=\(\dfrac{7}{5}+\dfrac{-1}{4}\) 

=\(\dfrac{23}{20}\) 

b) \(\left(\dfrac{-4}{3}+0,75\right):\dfrac{2017}{2018}+\left(1+\dfrac{1}{3}-75\%\right):\dfrac{2017}{2018}\) 

=\(\left(\dfrac{-4}{3}+0,75+1+\dfrac{1}{3}-75\%\right):\dfrac{2017}{2018}\) 

=\(\left[\left(\dfrac{-4}{3}+1+\dfrac{1}{3}\right)+\left(0,75-75\%\right)\right]:\dfrac{2017}{2018}\) 

=\(\left[0+0\right]:\dfrac{2017}{2018}\) 

=0\(:\dfrac{2017}{2018}\) 

=0

c)\(\left(2018-\dfrac{1}{3}-\dfrac{2}{4}-\dfrac{3}{5}-\dfrac{4}{6}-...-\dfrac{2018}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\)

=\(\left(1-\dfrac{1}{3}-1-\dfrac{2}{4}-1-\dfrac{3}{5}-1-\dfrac{4}{6}-...-1-\dfrac{2018}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) 

=\(\left(\dfrac{2}{3}-\dfrac{2}{4}-\dfrac{2}{5}-\dfrac{2}{6}-...-\dfrac{2}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left[2.\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-...-\dfrac{1}{2020}\right)\right]:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left\{2.\left[\dfrac{5}{5}.\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-...-\dfrac{1}{2020}\right)\right]\right\}:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left\{2.\left[5.\left(\dfrac{1}{15}-\dfrac{1}{20}-\dfrac{1}{25}-\dfrac{1}{30}-...-\dfrac{1}{10100}\right)\right]\right\}:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(10.\left(\dfrac{1}{15}-\dfrac{1}{20}-\dfrac{1}{25}-\dfrac{1}{30}-...-\dfrac{1}{10100}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =-10

\(A=1+2^1+2^2+...+2^{2017}\)

\(2A=2+2^2+2^3+...+2^{2018}\)

\(2A-A=2^{2018}-1hayA=2^{2018}-1\)

2; 3 tuong tu

1) A = 1 + 2 + 2² + 2³ + .... + 2²⁰¹⁸

2A = 2 + 2² + 2³ + 2⁴ + ..... + 2²⁰¹⁹

2A - A = ( 2 + 2² + 2³ + 2⁴ + ..... + 2²⁰¹⁹ ) - ( 1 + 2 + 2² + 2³ + .... + 2²⁰¹⁸ )

Vậy A = 2²⁰¹⁹ - 1

2) B = 1 + 3 + 3² + 3³ + ..... + 3²⁰¹⁸

3A = 3 + 3² + 3³ + ...... + 3²⁰¹⁹

3A - A = ( 3 + 3² + 3³ + ...... + 3²⁰¹⁹ ) - ( 1 + 3 + 3² + 3³ + ..... + 3²⁰¹⁸ )

2A = 3²⁰¹⁹ - 1

Vậy A = ( 3²⁰¹⁹ - 1 ) : 2

3) C = 1 + 4 + 4² + 4³ + ...... + 4²⁰¹⁸

4A = 4 + 4² + 4³ + ...... + 42019

4A - A = ( 4 + 4² + 4³ + ...... + 4²⁰¹⁹ ) - ( 1 + 4 + 4² + 4³ + ...... + 4²⁰¹⁸ )

3A = 4²⁰¹⁹ - 1

Vậy A = ( 4²⁰¹⁹ - 1 ) : 3

\(A=1+2+2^2+2^3+....+2^{2018}\)

\(2A=2+2^2+2^3+....+2^{2019}\)

\(A=2^{2019}-1\)

\(B=1+3+3^2+....+3^{2018}\)

\(3B=3+3^2+3^3+....+3^{2019}\)

\(2B=3^{2019}-1\)

\(B=\frac{2^{2019}-1}{2}\)

\(C=1+4+4^2+...+4^{2018}\)

\(\Rightarrow4B=4+4^2+4^3+...+4^{2019}\)

\(\Rightarrow3B=4^{2019}-1\)

\(\Rightarrow B=\frac{4^{2019}-1}{3}\)

Ta có công thức tổng quát sau: \(1-\frac{1}{1+2+\cdots+n}\)

\(=1-\frac{1}{\frac{n\left(n+1\right)}{2}}\)

\(=1-\frac{2}{n\left(n+1\right)}\)

\(=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}\)

\(=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)

Ta có: \(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\cdot\ldots\cdot\left(1-\frac{1}{1+2+3+\cdots+2018}\right)\)

\(=\frac{\left(2+2\right)\left(2-1\right)}{2\cdot\left(2+1\right)}\cdot\frac{\left(3+2\right)\left(3-1\right)}{3\cdot\left(3+1\right)}\cdot\ldots\cdot\frac{\left(2018+2\right)\left(2018-1\right)}{2018\cdot\left(2018+1\right)}\)

\(=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\ldots\cdot\frac{2020\cdot2017}{2018\cdot2019}\)

\(=\frac{4\cdot5\cdot\ldots\cdot2020}{3\cdot4\cdot\ldots\cdot2019}\cdot\frac{1\cdot2\cdot\ldots\cdot2017}{2\cdot3\cdot\ldots\cdot2018}=\frac{2020}{3}\cdot\frac{1}{2018}=\frac{1010}{1009\cdot3}=\frac{1010}{3027}\)