ai muôn kb vs mình ko

\(A=4\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\right)\)

\(=4\cdot\dfrac{2014}{2015}=\dfrac{8056}{2015}\)

\(\dfrac{4}{1.2}+\dfrac{4}{2.3}+...+\dfrac{4}{2021.2022}\\ =4\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2021.2022}\right)\\ =4\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\right)\\ =4\left(1-\dfrac{1}{2022}\right)\\ =4.\dfrac{2021}{2022}\\ =\dfrac{4042}{1011}\)

4/1.2 4/2.3 4/3.4 ... 4/2021.4/2022

= 1/4. (1/1- 1/2+ 1/2- 1/3+ 1/3- 1/4+...+1/2021- 1/2022)

=1/4. (1/1- 1/2022)= 1/4. (2022/2022- 1/2022)

= 1/4. 2021/2022

= 2021/8088

\(=4\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\right)=4\cdot\dfrac{2014}{2015}=\dfrac{8056}{2015}\)

\(\dfrac{4}{1.2}+\dfrac{4}{2.3}+...+\dfrac{4}{2014.2015}\\ =4\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2014.2015}\right)\\ =4\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\right)\\ =4\left(1-\dfrac{1}{2015}\right)\\ =4.\dfrac{2014}{2015}\\ =\dfrac{8056}{2015}\)

\(\dfrac{8056}{2015}\)

\(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{2019.2020}\)

\(\frac{1}{4}A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)

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

\(\frac{1}{4}A=1-\frac{1}{2020}=\frac{2019}{2020}\)

\(\Rightarrow A=\frac{2019}{2020}:\frac{1}{4}=\frac{2019}{505}\)

Vậy \(A=\frac{2019}{505}.\)

\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)

\(\Rightarrow2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\)

\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\)

\(2B=\frac{1}{1.2}-\frac{1}{99.100}=\frac{4949}{9900}\)

\(\Rightarrow B=\frac{4949}{9900}:2=\frac{4949}{19800}\)

Vậy \(B=\frac{4949}{19800}.\)

\(A=\frac{4}{1\cdot2}+\frac{4}{2\cdot3}+\frac{4}{3\cdot4}+...+\frac{4}{2019\cdot2020}\)

\(A=4\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}\right)\)

\(A=4\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)

\(A=4\left(1-\frac{1}{2019}\right)=4\cdot\frac{2018}{2019}\)

Đến đây tự tính

\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\)

\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{98\cdot99\cdot100}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{99\cdot100}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\)

Số hơi bị dữ nên tính nốt nhé

a) \(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+........+\frac{4}{2019.2020}\)

\(=4.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{2019.2020}\right)\)

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

\(=4.\left(1-\frac{1}{2020}\right)=4.\frac{2019}{2020}=\frac{2019}{505}\)

C=1.2+2.3+...+99.100

3C=1.2.3+2.3.3+...+99.100.3

3C=1.2(3-0)+2.3(4-1)+...+99.100(101-98)

C=99.100.101 phần 3

C=333 300

D=2²+4²+6²+333 300

D=4+16+36+333 300

D=20+36+333 300

D=56+333 300

D=333 356

Tích cho mình hai câu nhà, thank

a) \(C=1.2+2.3+3.4+...+99.100\)

\(\Rightarrow3C=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)

\(\Rightarrow3C=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)

\(\Rightarrow3C=99.100.101\)

\(\Rightarrow C=33.100.101\)

\(\Rightarrow C=333300\)

Bài 5:

a) Ta có: \(A=1\cdot2+2\cdot3+3\cdot4+...+9\cdot10\)

\(\Leftrightarrow3\cdot A=3\cdot\left(1\cdot2+2\cdot3+3\cdot4+...+9\cdot10\right)\)

\(\Leftrightarrow3A=1\cdot2\cdot\left(3-0\right)+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+9\cdot10\cdot\left(11-8\right)\)

\(\Leftrightarrow3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+8\cdot9\cdot10-8\cdot9\cdot10+9\cdot10\cdot11\)

\(\Leftrightarrow3\cdot A=9\cdot10\cdot11=90\cdot11=990\)

hay A=330

Vậy: A=330

a)đặt A = 1.2 + 3.4 + 4.5 +...+ 99.100

A=1.2+2.3+3.4+4.5+...+99.100

=>3A=1.2.3+2.3.3+3.4.3+4.5.3+...+99.100.3

=1.2.3+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+99.100.(101-98)

=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+99.100.101-98.99.100

=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+4.5.6-4.5.6+...+99.100.101 

=99.100.101=999900

=>A=999900:3=333300

Vậy A=333300 

b)D=2 +4 +6 +333 300

D=4+16+36+333 300

D=20+36+333 300 

D=56+333 300

D=333 356 

thank

a) C = 1.2 + 3.4 + 4.5 +...+ 99.100

\(\Rightarrow\) C = 1.2+2.3+3.4+4.5+...+99.100

\(\Rightarrow\)3C = 1.2.3+2.3.3+3.4.3+4.5.3+...+99.100.3

\(\Rightarrow\)3C = 1.2.3+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+99.100.(101-98)

 \(\Rightarrow\)3C = 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+99.100.101-98.99.100

 \(\Rightarrow\)3C = 1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+4.5.6-4.5.6+...+99.100.101 

\(\Rightarrow\)3C = 99.100.101

\(\Rightarrow C=33.100.101\)

\(\Rightarrow C=333300\)

b) D = 2 + 4 + 6 + 333 300

\(\Rightarrow\)D = 4 + 16 + 36 + 333 300

\(\Rightarrow\)D = 20 + 36 + 333 300 

\(\Rightarrow\)D = 56 + 333 300

\(\Rightarrow\)D = 333 356 

a, C = 1.2 + 2.3 + 3.4 + ... + 99.100

3C = 1.2.3 + 2.3.3 + 3.4.3 + .... + 99.100.3

3C = 1.2.3 + 2.3\((4-1)\)+ 3.4\((5-2)+...\)+ 99.100\((101-98)\)

3C = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100

3C = 99.100.101

C = 99.100.101 :3

C = 333300

Câu b tự làm

P/S : Hoq chắc :>