Lời giải:

$S=5(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2022.2023})$

$=5(\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2023-2022}{2022.2023})$

$=5(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2022}-\frac{1}{2023})$

$=5(\frac{1}{2}-\frac{1}{2023})=\frac{10105}{4046}$

\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)=2023x\)

\(\Rightarrow2022x+\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}+\dfrac{1}{2022}-\dfrac{1}{2023}\right)=2023x\)\(\Rightarrow2022x-2023x=-\left(1-\dfrac{1}{2023}\right)\)

\(\Rightarrow-x=-\dfrac{2022}{2023}\Leftrightarrow x=\dfrac{2022}{2023}\)

(x + 1/1.2) + (x + 1/2.3) + (x + 1/3.4) + ... + (x + 1/2022.2023) = 2023x

x + x + x + ... + x + 1/1.2 + 1/2.3 + ... + 1/2022.2023 = 2023x

2022x + 1 - 1/2 + 1/2 - 1/3 + ... + 1/2022 - 2023 = 2023x

2023x - 2022x = 1 - 1/2023

x = 2022/2023

\(S=\dfrac{2^2}{1.2}+\dfrac{2^2}{2.3}+\dfrac{2^2}{3.4}+...+\dfrac{2^2}{2022.2023}\)

\(S=2^2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)\)

\(S=2^2.\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)

\(S=2^2.\left(\dfrac{1}{1}-\dfrac{1}{2023}\right)\)

\(S=2^2.\dfrac{2022}{2023}\)

\(S=\dfrac{2^2.2022}{2023}=\dfrac{8088}{2023}\)

x-(1/1.2 + 1/2.3 + 1/3.4 + ...+ 1/2022.2023)= -2024/2023

x-(1-1/2 + 1/2-1/3 + 1/3-1/4 + ... + 1/2022-1/2023)=-2024/2023

x-(1-1/2023)=-2024/2023

x-2022/2023=-2024/2023

x = -2024/2023+2022/2023

x = -2/2023

Vậy x = -2/2023

A=1.2+2.3+3.4+.............+2019.2020

3A=1.2.3+2.3.3+3.4.3+........................+2019.2020.3

3A=1.2.3+2.3.(4-1)+3.4.(5-2)+..............+2019.2020.(2021-2018)

3A=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+.............-2018.2019.2020+2019.2020.2021

3A=2019.2020.2021

A=2019.2020.2021 / 3

A=2747468660

Vậy A=2747468660 .

🎀

4c=4.[1.2.3 + 2.3.4 + 3.4.5 + 4.5.6 + . . . . . . + 48.49.50] 
4c=1.2.3.4 +2.3.4.4 +3.4.5.4 +4.5.6.4 +.........+48.49.50.4 
4c=1.2.3.4 +2.3.4.(5-1) + 3.4.5.(6-2) +4.5.6(7-3)+....+ 48.49.50(51-47) 
4c=1.2.3.4 +2.3.4.5 --1.2.3.4 + 3.4.5.6-2.3.4.5 + 4.5.6.7-3.4.5.6+....+ 48.49.50.51-47.48.49.50   =48.49.50.51 
C=(48.49.50.51)/4     = 1499400

Câu b khúc cuối mik ko chc 😥

\(A=1.2+2.3+3,4+...+1999.2000\)

\(=>3A=1.2.3+2.3.3+3.4.3+...+1999.2000.3\)

\(=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+.....+1999.2000.\left(2001-1998\right)\)

\(=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+1999.2000.2001-1998.1999.2000\)

\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+1999.2000.2001-1998.1999.2000\)

\(=1999.2000.2001\)

\(=>A=\frac{1999.2000.2001}{3}=......\) (bn dùng máy tính)

b,xem lại chỗ 3.2

c,tính 4C , biến đổi tương tự câu a
 

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