\(\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{97\cdot99}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{99}\right)=\dfrac{1}{2}\cdot\dfrac{32}{99}=\dfrac{16}{99}\)

\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{95.97}+\dfrac{1}{97.99}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{99}\)

\(=\dfrac{1}{3}-\dfrac{1}{99}\)

\(=\dfrac{33}{99}-\dfrac{1}{99}\)

\(=\dfrac{32}{99}\)

\(A=\dfrac{12}{\left(2.4\right)^2}+\dfrac{20}{\left(4.6\right)^2}+...+\dfrac{388}{\left(96.98\right)^2}+\dfrac{396}{\left(98.100\right)^2}\)

\(A=\dfrac{1}{2^2}-\dfrac{1}{4^2}+\dfrac{1}{4^2}-\dfrac{1}{6^2}+...+\dfrac{1}{96^2}-\dfrac{1}{98^2}+\dfrac{1}{98^2}-\dfrac{1}{100^2}\)

\(A=\dfrac{1}{2^2}-\dfrac{1}{100^2}\)

\(A=\dfrac{1}{4}-\dfrac{1}{100^2}< \dfrac{1}{4}\)

\(\Rightarrow A< \dfrac{1}{4}\)

Gọi tổng \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\) là A.

Theo bài ra, ta có:

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\\ A=\dfrac{1}{2\cdot2}+\dfrac{1}{3\cdot3}+\dfrac{1}{4\cdot4}+...+\dfrac{1}{100\cdot100}\\ \Rightarrow A< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\\ \Rightarrow A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ \Rightarrow A< \dfrac{1}{1}-\dfrac{1}{100}\\ \Rightarrow A< \dfrac{99}{100}< 1\)

Vậy \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)

Cho em sửa đề ạ * 1-3+5-7+...+2017-2019+2021

D=-2022 nhé chominhf 1 tích nhé

\(2\cdot x=16\\ x=16:2\\ x=8\)

Vậy \(x=8\)

2nhân x =16

x=16:2

x=8

vậy x=8

bừng o bạn nhé cho mình 1 tịh

C= 1 - 2 - 3 + 4 + 5 - 6 - 7 + ... + 2021 - 2022 -2023 + 2024

C= (1-2-3+4) + (5-6-7+8) + ... + (2021-2022-2023+2024)

C= 0 + 0 + ... + 0

C = 0

Vậy C=0

Giá vẻ trẻ em là:

\(2800000\cdot75\%=2100000\left(đồng\right)\)

Số tiền nhà bạn An phải trả là:

\(\left(2\cdot2800000+2\cdot2100000\right)\left(1-20\%\right)=0,8\cdot9800000=7840000\left(đồng\right)\)

\(\dfrac{6}{7}\cdot\dfrac{8}{13}+\dfrac{6}{13}\cdot\dfrac{9}{7}-\dfrac{4}{13}\cdot\dfrac{6}{7}\)

\(=\dfrac{6}{7}\left(\dfrac{8}{13}+\dfrac{9}{13}-\dfrac{4}{13}\right)\)

\(=\dfrac{6}{7}\cdot\dfrac{13}{13}=\dfrac{6}{7}\)

   \(\dfrac{6}{7}\).\(\dfrac{8}{13}\) + \(\dfrac{6}{13}\).\(\dfrac{9}{7}\) - \(\dfrac{4}{13}\).\(\dfrac{6}{7}\)

=  \(\dfrac{6}{7}\).\(\dfrac{8}{13}\) + \(\dfrac{6}{7}\).\(\dfrac{9}{13}\) - \(\dfrac{4}{13}\).\(\dfrac{6}{7}\)

= \(\dfrac{6}{7}\).(\(\dfrac{8}{13}\) + \(\dfrac{9}{13}\) - \(\dfrac{4}{13}\))

= \(\dfrac{6}{7}\).1

= \(\dfrac{6}{7}\)

Quãng đường còn lại ứng với phân số là:

           1  - \(\dfrac{2}{3}\) = \(\dfrac{1}{3}\) (quãng đường)

Quãng đường còn lại dài số ki-lô-mét là:

          42 x \(\dfrac{1}{3}\) = 14 (km)

Kết luận: Quãng đường còn lại dài 14km