\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(A=1-\frac{1}{2020}\)

\(A=\frac{2019}{2020}\)

\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)

\(2B=\frac{2}{1.3}+\frac{2}{3.5}=\frac{2}{5.7}+...+\frac{2}{2017.2019}\)

\(2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}=\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)

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

\(2B=\frac{2018}{2019}\)

\(B=\frac{2018}{2019}:2=\frac{1009}{2019}\)

\(C=3^0+3^1+3^2+...+3^{100}\)

\(3C=3^1+3^2+3^3+...+3^{101}\)

\(3C-C=\left(3^1+3^2+3^3+...+3^{101}\right)-\left(1+3^1+3^2+...+3^{100}\right)\)

\(2C=3^{101}-1\)

\(C=\frac{3^{101}-1}{2}\)

Ta có

 1 x 2 x 3 x .... x 2019 x 2020 chữ số tận cùng là 0

1 x 3 x 5 x ... x 2017 x 2019 chữ số tận cùng là 5

  Vậy A = 1 x 2 x 3 x .... x 2019 x 2020 - 1 x 3 x 5 x .... x 2017 x 2019 chữ số tận cùng sẽ là 5

chữ số tận cùng sẽ là 5

1. Tự làm

2. Ta có: \(x_1+x_2+x_3+...+x_{2017}+x_{2018}+x_{2019}+x_{2020}=0\)

=> \(\left(x_1+x_2+x_3\right)+\left(x_4+x_5+x_6\right)+....+\left(x_{2017}+x_{2018}+x_{2019}\right)+x_{2020}=0\)

=> \(3+3+....+3+x_{2020}=0\) (gồm 673 chữ số 3 vì x₁ + .... + x₂₀₁₉ gồm 2019 hạng tử gộp lại mỗi cặp 3 hạng tử)

=> \(3.673+x_{2020}=0\)

=> \(2019+x_{2020}=0\)

=> \(x_{2020}=-2019\)

3. a) 3(x - 1) - (x - 5) = -18

=> 3x - 3 - x + 5 = -18

=> 2x + 2 = -18

=> 2x  = -18 - 2

=> 2x = -20

=> x = -20 : 2

=> x = 10

b ) x + (x + 1) + (x + 2) + ... + (x + 2019) = 0

=> (x + x  + ... + x) + (1 + 2 + ...  + 2019) = 0

=> 2020x + (2019 + 1).[(2019 - 1) : 1 + 1] : 2 = 0

=> 2020x + 2020. 2019 : 2 = 0

=> 2020x + 2039190 = 0

=> 2020x = -2039190

=> x = -2039190 : 2020

=> x = -10095 

(xem lại đề)

c) Ta có: 3x + 23 = 3(x + 4) + 11

Do 3(x + 4) \(⋮\)4 => 11 \(⋮\)x + 4

=> x + 4 \(\in\)Ư(11) = {1; -1; 11; -11}

Với: +) x + 4 = 1 => x = 1 - 4 = -3

+) x + 4 = -1 => x = -1 - 4 = -5

+) x + 4 = 11 => x = 11 - 4 = 7

+) x + 4 = -11 => x = -11 - 4 = -15

4a) Ta có: 22x - y = 21x + x - y = 21 + (x - y)

Do 21x \(⋮\)7; x - y \(⋮\)7

=> 22x - y \(⋮\)7

b) 8x + 20y = 7x + 21y + x - y = 7(x + 3y) + (x - y)

Do : 7(x + 3y) \(⋮\)7; x - y \(⋮\)7

=> 8x + 20y \(⋮\)7

c) 11x + 10y = 14x + 7y - 3x + 3y = 7(2x + y) - 3(x - y)

Do: 7(2x + y) \(⋮\)7; 3(x - y) \(⋮\)7

=> 11x + 10y \(⋮\)7

\(A=\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{2017\cdot2019}\)

\(A=\frac{1}{2}\left(\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{2017\cdot2019}\right)\)

\(A=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)

\(A=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{2019}\right)\)

\(A=\frac{1}{2}\cdot\frac{2014}{10095}\)

\(A=\frac{1007}{10095}\)

\(A=\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{2015\cdot2017}+\frac{1}{2017\cdot2019}\)

\(2A=\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{2015\cdot2017}+\frac{2}{2017\cdot2019}\)

\(2A=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2015}-\frac{1}{2017}+\frac{1}{2017}-\frac{1}{2019}\)

\(2A=\frac{1}{5}-\frac{1}{2019}\)

\(2A=\frac{2014}{100095}\)

\(A=\frac{2014}{10095}:2=\)TỰ TÍNH

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2017.2019}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)

\(=1-\frac{1}{2019}\)

\(=\frac{2018}{2019}\)

=1-1/3+1/3-1/5+....+1/2017-1/2019

=1-1/2019=2018/2019

t i ck nha

a)x=0

a) \(\frac{x}{2}+\frac{x}{3}+\frac{x}{4}+\frac{x}{5}=0\)

\(\frac{77x}{60}=0\)

\(77x=0.60\)

\(77x=0\)

\(x=0\)

\(\frac{a+1}{2019}+\frac{a+2}{2018}=\frac{a+3}{2017}+\frac{a+4}{2016}\Leftrightarrow\frac{a+2020}{2019}+\frac{a+2020}{2018}=\frac{a+2020}{2017}+\frac{a+2020}{2016}\) 

\(\left(a+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)=0\Rightarrow a+2020=0\Leftrightarrow a=-2020\)