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 !!!
K
Khách
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Những câu hỏi liên quan
5 tháng 11 2017
\(S=\dfrac{1}{2}-\dfrac{1}{3.7}-\dfrac{1}{7.11}-...........-\dfrac{1}{23.27}\)
\(=\dfrac{1}{2}-\left(\dfrac{1}{3.7}+\dfrac{1}{7.11}+..........+\dfrac{1}{23.27}\right)\)
\(=\dfrac{1}{2}-\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+.......+\dfrac{1}{23}-\dfrac{1}{27}\right)\)
\(=\dfrac{1}{2}-\left(\dfrac{1}{3}-\dfrac{1}{27}\right)\)
\(=\dfrac{1}{2}-\dfrac{8}{27}\)
\(=\dfrac{11}{54}\)
5 tháng 11 2017
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 
7 tháng 8 2018
\(\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{23.27}\)
= \(4.\left(\text{}\text{}\text{}\text{}\text{}\text{}\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{23.27}\right)\)
=\(1.\left(\dfrac{1}{3.7}+\dfrac{1}{7.11}+...+\dfrac{1}{23.27}\right)\)
= \(1.\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{23}-\dfrac{1}{27}\right)\)
=\(1.\left(\dfrac{1}{3}-\dfrac{1}{27}\right)\)
=\(1.\left(\dfrac{9}{27}-\dfrac{1}{27}\right)\)
= \(1.\dfrac{8}{27}\)
= \(\dfrac{8}{27}\)
24 tháng 10 2017
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
25 tháng 3 2022
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2021.2022}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)
\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)
\(=\dfrac{1}{3}-\dfrac{1}{111}=\dfrac{12}{37}\)
KV
11 tháng 9 2023
a) 1/(5.7) + 1/(7.9) + ... + 1/(2011.2013)
= 1/2.(1/5 - 1/7 + 1/7 - 1/9 + ... + 1/2011 - 1/2013)
= 1/2.(1/5 - 1/2013)
= 1/2 . 2008/10065
= 1004/10065
b) 1/(7.11) + 1/(11.15) +1/(15.19) + ... + 1/(2019.2023)
= 1/4.(1/7 - 1/11 + 1/11 - 1/15 + 1/15 - 1/19 + ... + 1/2019 - 1/2023)
= 1/4.(1/7 - 1/2023)
= 1/4 . 288/2023
= 72/2023
23 tháng 3 2023
\(A=\dfrac{-5}{3}\cdot\dfrac{11}{2}\cdot\dfrac{4}{3}=\dfrac{-20\cdot11}{2\cdot9}=\dfrac{-110}{9}\)
\(B=\dfrac{2}{4}\left(\dfrac{4}{11\cdot15}+\dfrac{4}{15\cdot19}+...+\dfrac{4}{51\cdot55}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+...+\dfrac{1}{51}-\dfrac{1}{55}\right)\)
=1/2*4/55
=2/55
KV
6 tháng 11 2023
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
AH
Akai Haruma
Giáo viên
6 tháng 11 2023
Lời giải:
$A=5(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{2019.2023})$
$4A=5(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{2019.2023})$
$=5(\frac{7-3}{3.7}+\frac{11-7}{7.11}+\frac{15-11}{11.15}+...+\frac{2023-2019}{2019.2023})$
$=5(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+....+\frac{1}{2019}-\frac{1}{2023})$
$=5(\frac{1}{3}-\frac{1}{2023})=\frac{2020}{6069}$
$\Rightarrow A=\frac{2020}{6069}:4=\frac{505}{6069}$
LG
2
XO
28 tháng 7 2020
Ta có : \(\frac{1}{2}-\frac{1}{3.7}-\frac{1}{7.11}-\frac{1}{11.15}-\frac{1}{15.19}-\frac{1}{19.23}-\frac{1}{23.27}\)
\(=\frac{1}{2}-\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\frac{1}{15.19}+\frac{1}{19.23}+\frac{1}{23.27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{27}\right)=\frac{1}{2}-\frac{1}{4}.\frac{8}{27}=\frac{1}{2}-\frac{2}{27}=\frac{23}{54}\)
28 tháng 7 2020
Trả lời:
\(\frac{1}{2}-\frac{1}{3.7}-\frac{1}{7.11}-\frac{1}{11.15}-\frac{1}{15.19}-\frac{1}{19.23}-\frac{1}{23.27}\)
\(=\frac{1}{2}-\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\frac{1}{15.19}+\frac{1}{19.23}+\frac{1}{23.27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}.\frac{8}{27}\)
\(=\frac{1}{2}-\frac{2}{27}\)
\(=\frac{23}{54}\)
Học tốt
AH
Akai Haruma
Giáo viên
9 tháng 5 2021
Lời giải:
\(B=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+....+\frac{2021}{4^{2021}}\)
\(4B=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2021}{4^{2020}}\)
\(4B-B=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2020}}-\frac{2021}{4^{2021}}\)
\(3B=1+\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2020}}-\frac{2021}{4^{2021}}\)
\(12B=4+1+\frac{1}{4}+...+\frac{1}{4^{2019}}-\frac{2021}{4^{2020}}\)
\(9B=4-\frac{6067}{4^{2021}}<4\Rightarrow B< \frac{4}{9}< \frac{1}{2}\)
bạn tham khảo nha, cách làm như vậy đó
Câu hỏi của Nguyễn Thị Mai Ca - Toán lớp 7 - Học toán với OnlineMath
ban kia lam dung roi do
k tui nha
thanks