Bài 1: Tính nhanh: \(\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+\dfrac{4}{15.19}+\dfrac{4}{19.23}+\dfrac{4}{23.27}\)
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 !!!
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
Tính :
a) 4/3.7 - 4/7.11 + 4/11.15 - 4/15.19 + 4/19.23 - 4/23.27
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
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{2021.2022}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+......+\dfrac{4}{107.111}\)
\(S=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+.....+\dfrac{1}{60}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2021.2022}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)
\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)
\(=\dfrac{1}{3}-\dfrac{1}{111}=\dfrac{12}{37}\)
\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)
\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{23.27}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\)
\(=\frac{1}{3}-\frac{1}{27}+0+0+0+0\)
\(=\frac{8}{27}\)
Ta có : \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.....+\frac{1}{23}-\frac{1}{27}\)
\(=\frac{1}{3}-\frac{1}{27}\)
\(=\frac{8}{27}\)
\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)
\(=\frac{7-3}{3.7}+\frac{11-7}{7.11}+.....+\frac{27-23}{23.27}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{23}-\frac{1}{27}\)
\(=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
Tìm x, biết: \(\dfrac{4}{3.7}\)+\(\dfrac{4}{7.11}\)+\(\dfrac{4}{11.15}\)+...+\(\dfrac{4}{\left(3x-1\right)\left(3x+3\right)}\)=\(\dfrac{3}{10}\)
\(\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{\left(3x-1\right)\left(3x+3\right)}=\dfrac{3}{10}\) \(\Rightarrow\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{15}+...+\dfrac{1}{\left(3x-1\right)}-\dfrac{1}{\left(3x+3\right)}=\dfrac{3}{10}\)\(\Rightarrow\dfrac{1}{3}-0-0-...-0-\dfrac{1}{\left(3x+3\right)}=\dfrac{3}{10}\)(cộng số đối)
\(\Rightarrow\dfrac{1}{3}-\dfrac{1}{\left(3x+3\right)}=\dfrac{3}{10}\)
\(\Rightarrow\dfrac{1}{\left(3x+3\right)}=\dfrac{1}{3}-\dfrac{3}{10}\)
\(\Rightarrow\dfrac{1}{\left(3x+3\right)}=\dfrac{1}{30}\)
\(\Rightarrow3x+3=30\)
\(\Rightarrow x=\left(30-3\right)+3=9\)
Vậy x=9
Tính :
A=\(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+......+\dfrac{1}{120}\)
B=\(\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+......+\dfrac{4}{107.111}\)
C=\(\dfrac{7}{10.11}+\dfrac{7}{11.12}+\dfrac{7}{12.13}+......+\dfrac{7}{69.70}\)
D=\(\dfrac{15}{90.94}+\dfrac{15}{94.98}+\dfrac{15}{98.102}+......+\dfrac{15}{146.150}\)
\(A=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)
\(A=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)
\(A=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{15.16}\)
\(A=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)
\(A=2.\dfrac{3}{16}\)
\(A=\dfrac{3}{8}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)
\(B=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)
\(B=\dfrac{1}{3}-\dfrac{1}{111}\)
\(B=\dfrac{12}{37}\)
\(C=\dfrac{7}{10.11}+\dfrac{7}{11.12}+\dfrac{7}{12.13}+...+\dfrac{7}{69.70}\)
\(C=7\left(\dfrac{1}{10.11}+\dfrac{1}{11.12}+\dfrac{1}{12.13}+...+\dfrac{7}{69.70}\right)\)
\(C=7\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)
\(C=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)\)
\(C=7.\dfrac{3}{35}\)
\(C=\dfrac{3}{5}\)
1.6,cho A=\(\dfrac{4}{15.19}+\dfrac{4}{19.23}+...+\dfrac{4}{399.403}\)CMR:\(\dfrac{16}{81}< A< \dfrac{16}{80}\)