tính:
A= 1.1+2.3+3.4+....+2022.2023
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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
\(S=\dfrac{2^2}{1.2}+\dfrac{2^2}{2.3}+\dfrac{2^2}{3.4}+...+\dfrac{2^2}{2022.2023}\)
\(S=2^2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)\)
\(S=2^2.\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)
\(S=2^2.\left(\dfrac{1}{1}-\dfrac{1}{2023}\right)\)
\(S=2^2.\dfrac{2022}{2023}\)
\(S=\dfrac{2^2.2022}{2023}=\dfrac{8088}{2023}\)
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 .
🎀
\(A=1.2+2.3+3,4+...+1999.2000\)
\(=>3A=1.2.3+2.3.3+3.4.3+...+1999.2000.3\)
\(=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+.....+1999.2000.\left(2001-1998\right)\)
\(=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+1999.2000.2001-1998.1999.2000\)
\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+1999.2000.2001-1998.1999.2000\)
\(=1999.2000.2001\)
\(=>A=\frac{1999.2000.2001}{3}=......\) (bn dùng máy tính)
b,xem lại chỗ 3.2
c,tính 4C , biến đổi tương tự câu a
Lời giải:
$A=1+2.3+3.4+4.5+...+2022.2023$
$3A=3+2.3(4-1)+3.4(5-2)+4.5(6-3)+....+2022.2023(2024-2021)$
$=3+2.3.4+3.4.5+4.5.6+...+2022.2023.2024-(1.2.3+2.3.4+3.4.5+...+2021.2022.2023)$
$=3+2022.2023.2024-1.2.3=2022.2023.2024-3$
$\Rightarrow A=2759728047$