a) \(\dfrac{1}{5.7}\)+\(\dfrac{1}{7.9}\)+...+\(\dfrac{1}{2011.2013}\)
b) \(\dfrac{1}{7.11}\)+\(\dfrac{1}{11.15}\)+\(\dfrac{1}{15.19}\)+...\(\dfrac{1}{2019.2023}\)
Bài 1: Tính nhanh: \(\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+\dfrac{4}{15.19}+\dfrac{4}{19.23}+\dfrac{4}{23.27}\)
\(\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{23.27}\)
= \(4.\left(\text{}\text{}\text{}\text{}\text{}\text{}\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{23.27}\right)\)
=\(1.\left(\dfrac{1}{3.7}+\dfrac{1}{7.11}+...+\dfrac{1}{23.27}\right)\)
= \(1.\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{23}-\dfrac{1}{27}\right)\)
=\(1.\left(\dfrac{1}{3}-\dfrac{1}{27}\right)\)
=\(1.\left(\dfrac{9}{27}-\dfrac{1}{27}\right)\)
= \(1.\dfrac{8}{27}\)
= \(\dfrac{8}{27}\)
a, D= \(\left(2\dfrac{2}{15}.\dfrac{9}{17}.\dfrac{3}{32}\right):\left(-\dfrac{3}{17}\right)\)
b, \(\dfrac{1}{2}-\dfrac{1}{3.7}-\dfrac{1}{7.11}-\dfrac{1}{11.15}-\dfrac{1}{15.19}-\dfrac{1}{19.23}-\dfrac{1}{23.27}\)
c, 1- \(\dfrac{1}{5.10}-\dfrac{1}{10.15}-\dfrac{1}{15.20}-.......-\dfrac{1}{95.100}\)
a) \(D=\left(2\dfrac{2}{15}\times\dfrac{9}{17}\times\dfrac{3}{32}\right)\div\left(-\dfrac{3}{17}\right)\)
\(D=\dfrac{32}{15}\times\dfrac{9}{17}\times\dfrac{3}{32}\times\dfrac{-17}{3}\)
\(D=\dfrac{-3}{5}\)
b) \(\dfrac{1}{2}-\dfrac{1}{3\times7}-\dfrac{1}{7\times11}-\dfrac{1}{11\times15}-\dfrac{1}{15\times19}-\dfrac{1}{19\times23}-\dfrac{1}{23\times27}\)
\(=\dfrac{1}{2}-\left(\dfrac{1}{3\times7}+\dfrac{1}{7\times11}+\dfrac{1}{11\times15}+\dfrac{1}{15\times19}+\dfrac{1}{19\times23}+\dfrac{1}{23\times25}\right)\)
\(=\dfrac{1}{2}-\left[\dfrac{1}{4}\left(\dfrac{4}{3\times7}+\dfrac{4}{7\times11}+\dfrac{4}{11\times15}+\dfrac{4}{15\times19}+\dfrac{4}{19\times23}+\dfrac{4}{23\times27}\right)\right]\)
\(=\dfrac{1}{2}-\left[\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{27}\right)\right]\)
\(=\dfrac{1}{2}-\left[\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{27}\right)\right]\)
\(=\dfrac{1}{2}-\left[\dfrac{1}{4}\left(\dfrac{9-1}{27}\right)\right]\)
\(=\dfrac{1}{2}-\dfrac{1}{4}\times\dfrac{8}{27}\)
\(=\dfrac{1}{2}-\dfrac{2}{27}\)
\(=.....\)
Đó đến đây bn tự lm nốt. Câu c bn lm tương tự.
Mình cho bn dạng này, nếu bn chưa biết (để lm câu c)
\(\dfrac{x}{y\left(y+x\right)}=\dfrac{x}{y}-\dfrac{x}{y+x}\)
Chúc bn học tốt
So sánh tổng S với 251
S = \(\dfrac{1}{2}-\dfrac{1}{3.7}-\dfrac{1}{7.11}-\dfrac{1}{11.15}-\dfrac{1}{15.19}-\dfrac{1}{19.23}-\dfrac{1}{23.27}\)
Mai mk thi r cho mình xem cách làm bài này nhé. Giúp mình với. HELP ME !!!
tính tích A>B biết:
\(A=\left(-\dfrac{5}{3}\right).5\dfrac{1}{2}.\left(\dfrac{1}{3}+1\right)\)
\(B=\dfrac{2}{11.15}+\dfrac{2}{15.19}+\dfrac{2}{19.23}+...+\dfrac{2}{51.55}\)
\(A=\dfrac{-5}{3}\cdot\dfrac{11}{2}\cdot\dfrac{4}{3}=\dfrac{-20\cdot11}{2\cdot9}=\dfrac{-110}{9}\)
\(B=\dfrac{2}{4}\left(\dfrac{4}{11\cdot15}+\dfrac{4}{15\cdot19}+...+\dfrac{4}{51\cdot55}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+...+\dfrac{1}{51}-\dfrac{1}{55}\right)\)
=1/2*4/55
=2/55
So sánh tổng S với 251
S = \(\dfrac{1}{2}-\dfrac{1}{3.7}-\dfrac{1}{7.11}-\dfrac{1}{11.15}-\dfrac{1}{15.19}-\dfrac{1}{19.23}-\dfrac{1}{23.27}\)
Mai mk thi r cho mình xem cách làm bài này nhé. Giúp mình với. HELP ME !!!
\(S=\dfrac{1}{2}-\dfrac{1}{3.7}-\dfrac{1}{7.11}-...........-\dfrac{1}{23.27}\)
\(=\dfrac{1}{2}-\left(\dfrac{1}{3.7}+\dfrac{1}{7.11}+..........+\dfrac{1}{23.27}\right)\)
\(=\dfrac{1}{2}-\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+.......+\dfrac{1}{23}-\dfrac{1}{27}\right)\)
\(=\dfrac{1}{2}-\left(\dfrac{1}{3}-\dfrac{1}{27}\right)\)
\(=\dfrac{1}{2}-\dfrac{8}{27}\)
\(=\dfrac{11}{54}\)
Bạn xem lại đề bài đi chứ thế này thì cần j phải so sánh nx
Này nhé: đã có \(\dfrac{1}{2}=2^{-1}\) mà \(2^{-1}< 2^{51}\) là điều quá rõ rồi
Đã thế lại còn trừ liên hoàn từ... (đấy nói chung là phần sau) thì rõ ràng hiển nhiên là \(S< 2^{51}\) còn cái j nx
Chúc bn học tốt
tính các tổng sau bằng cách hợp lí
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{2021.2022}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+......+\dfrac{4}{107.111}\)
\(S=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+.....+\dfrac{1}{60}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2021.2022}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)
\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)
\(=\dfrac{1}{3}-\dfrac{1}{111}=\dfrac{12}{37}\)
Tính tổng A=\(\dfrac{5}{3.7}+\dfrac{5}{7.11}+\dfrac{5}{11.15}+...+\dfrac{5}{2019.2023}\)
các bạn làm đầy đủ các bước đừng làm tắt giúp mình nhé.Mình cảm ơn ạ!
A = 5/(3.7) + 5/(7.11) + 5/(11.15) + ... + 5/(2019.2023)
= 5/4 . (1/3 - 1/7 + 1/7 - 1/11 + 1/11 - 1/15 + ... + 1/2019 - 1/2023)
= 5/4 . (1/3 - 1/2023)
= 5/4 . 2020/6069
= 2525/6069
Lời giải:
$A=5(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{2019.2023})$
$4A=5(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{2019.2023})$
$=5(\frac{7-3}{3.7}+\frac{11-7}{7.11}+\frac{15-11}{11.15}+...+\frac{2023-2019}{2019.2023})$
$=5(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+....+\frac{1}{2019}-\frac{1}{2023})$
$=5(\frac{1}{3}-\frac{1}{2023})=\frac{2020}{6069}$
$\Rightarrow A=\frac{2020}{6069}:4=\frac{505}{6069}$
A = \(\dfrac{5}{3.7}\) + \(\dfrac{5}{7.11}\) + \(\dfrac{5}{11.15}\)+...+ \(\dfrac{5}{2019.2023}\)
A = 5.(\(\dfrac{1}{3.7}\) + \(\dfrac{1}{7.11}\) + \(\dfrac{1}{11.15}\) + ... + \(\dfrac{1}{2019.2023}\))
A = \(\dfrac{5}{4}\).( \(\dfrac{4}{3.7}\) + \(\dfrac{4}{7.11}\) + \(\dfrac{4}{11.15}\) + ... + \(\dfrac{4}{2019.2023}\))
A = \(\dfrac{5}{4}\).( \(\dfrac{1}{3}\) - \(\dfrac{1}{7}\) +\(\dfrac{1}{7}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{15}\) + ... + \(\dfrac{1}{2019}\) - \(\dfrac{1}{2023}\))
A = \(\dfrac{5}{4}\). ( \(\dfrac{1}{3}\) - \(\dfrac{1}{2023}\))
A = \(\dfrac{5}{4}\) . \(\dfrac{2020}{6069}\)
A = \(\dfrac{2525}{6069}\)
B=\(\dfrac{-1}{3}\)+ \(\dfrac{-1}{3.5}\)+ \(\dfrac{-1}{5.7}\)+ \(\dfrac{-1}{7.9}\) +... + \(\dfrac{-1}{99.101}\)
\(B=-\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{-1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{100}{101}=-\dfrac{50}{101}\)
a/ \(A=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{99.100}\)
b/ \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{20}}< 1\)
c/ \(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\)
d/ \(A=\dfrac{2015}{2016}+\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2015}>4\)