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NT
2
6 tháng 1 2022
\(S=\dfrac{1}{5\times9}+\dfrac{1}{9\times13}+...+\dfrac{1}{41\times45}\)
\(\Rightarrow4S=\dfrac{4}{5\times9}+\dfrac{4}{9\times13}+...+\dfrac{4}{41\times45}\)
\(\Rightarrow4S=\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}\)
\(\Rightarrow4S=\dfrac{1}{5}-\dfrac{1}{45}\)
\(\Rightarrow4S=\dfrac{8}{45}\)
\(\Rightarrow S=\dfrac{2}{45}\)
6 tháng 1 2022
\(=\dfrac{1}{4}\left(\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{8}{45}=\dfrac{2}{45}\)
PP
2
NH
6 tháng 6 2019
\(\frac{11}{1.4}+\frac{11}{4.7}+...+\frac{11}{100.103}\)
\(=\frac{11}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(=\frac{11}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=\frac{11}{3}\left(1-\frac{1}{103}\right)\)
Tự tính
XO
6 tháng 6 2019
\(\frac{11}{1.4}+\frac{11}{4.7}+...+\frac{11}{100.103}\)
= \(\frac{11}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
= \(\frac{11}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
= \(\frac{11}{3}.\left(1-\frac{1}{103}\right)\)
= \(\frac{11}{3}.\frac{102}{103}\)
= \(\frac{374}{103}\)
KV
31 tháng 10 2023
3/(1×5) + 3/(5×9) + 3/(9×13) + 3/(13×17) + 3/(17×21)
= 3/4 × (1 - 1/5 + 1/5 - 1/9 + 1/9 - 1/13 + 1/13 - 1/17 + 1/17 - 1/21)
= 3/4 × (1 - 1/21)
= 3/4 × 20/21
= 5/7
TN
3
27 tháng 6 2019
4/5x9 + 4/9x13 + 4/13x17 + 4/ 17x21 + 4/ 21x25=27.911628241
27 tháng 6 2019
Ta có:
\(\frac{4}{5\times9}+\frac{4}{9\times13}+\frac{4}{13\times17}+\frac{4}{17\times21}+\frac{4}{21\times25}\)
= \(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\)
= \(\frac{1}{5}-\left(\frac{1}{9}+\frac{1}{9}\right)-\left(\frac{1}{13}-\frac{1}{13}\right)-\left(\frac{1}{17}-\frac{1}{17}\right)-\left(\frac{1}{21}-\frac{1}{21}\right)-\frac{1}{25}\)
= \(\frac{1}{5}-\frac{1}{25}\)
= \(\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
24 tháng 11 2023
\(\dfrac{4\times x}{1\times5}\) + \(\dfrac{4\times x}{5\times9}\) + \(\dfrac{4\times x}{9\times13}\) + \(\dfrac{4\times x}{13\times17}\) = 16
\(x\times\left(\dfrac{4}{1\times5}+\dfrac{4}{5\times9}+\dfrac{4}{9\times13}+\dfrac{4}{13\times17}\right)\) = 16
\(x\) \(\times\) (\(\dfrac{1}{1}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) - \(\dfrac{1}{13}\) + \(\dfrac{1}{13}\) - \(\dfrac{1}{17}\)) = 16
\(x\) \(\times\) ( \(\dfrac{1}{1}\) - \(\dfrac{1}{17}\)) = 16
\(x\) \(\times\) \(\dfrac{16}{17}\) = 16
\(x\) = 16 : \(\dfrac{16}{17}\)
\(x\) = 17
15 tháng 10 2021
\(\Rightarrow\dfrac{7}{x}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{29}{45}\left(x\ne0\right)\\ \Rightarrow\dfrac{7}{x}+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{29}{45}\\ \Rightarrow\dfrac{7}{x}+\dfrac{8}{45}=\dfrac{29}{45}\\ \Rightarrow\dfrac{7}{x}=\dfrac{21}{45}\Rightarrow\dfrac{21}{3x}=\dfrac{21}{45}\Rightarrow3x=45\\ \Rightarrow x=15\)
PT
4
23 tháng 5 2023
Đáp án + Giải thích các bước giải:
7/1×5+7/5×9+7/9×13+7/13×17+7/17×21
=7/4×(4/1×5+4/5×9+4/9×13+4/13×17+4/17×21)
=7/4×(1−1/5+1/5−1/9+1/9−1/13+1/13−1/17+1/17−1/21)
=7/4×(1+0+0+0+0−1/21)
=7/4×(1−1/21)
=7/4×2021
=5/3
nhớ tick nha
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 5 2023
\(\dfrac{7}{1\times5}+\dfrac{7}{5\times9}+\dfrac{7}{9\times13}+\dfrac{7}{13\times17}\)\(+\) \(\dfrac{7}{17\times21}\)
= \(\dfrac{7}{4}\times\)( \(\dfrac{4}{1\times5}\)\(+\) \(\dfrac{4}{5\times9}\)\(+\) \(\dfrac{4}{9\times13}\)\(+\)\(\dfrac{4}{13\times17}\)\(+\)\(\dfrac{4}{17\times21}\))
= 7 \(\times\)(\(\dfrac{1}{1}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{17}-\dfrac{1}{21}\))
= 7\(\times\)(\(\dfrac{1}{1}-\dfrac{1}{21}\))
= \(\dfrac{7}{4}\)\(\times\) \(\dfrac{20}{21}\)
= \(\dfrac{5}{3}\)
A
22 tháng 7 2018
a. 5/12+(7/59+7/12)
=5/12+497/708
=66/59
b.(7/30+5/16)+(1/16-7/30)
=131/240+(-41/240)
=3/8
KT
22 tháng 7 2018
a) \(\frac{5}{12}+\left(\frac{7}{59}+\frac{7}{12}\right)\)
\(=\frac{5}{12}+\frac{7}{59}+\frac{7}{12}\)
\(=\left(\frac{5}{12}+\frac{7}{12}\right)+\frac{7}{59}\)
\(=1\frac{7}{59}\)
b) \(\left(\frac{7}{30}+\frac{5}{16}\right)+\left(\frac{1}{16}-\frac{7}{30}\right)\)
\(=\frac{7}{30}+\frac{5}{16}+\frac{1}{16}-\frac{7}{30}\)
\(=\frac{6}{16}=\frac{3}{8}\)
c) \(\frac{3}{5.9}+\frac{3}{9.13}+\frac{3}{13.17}+...+\frac{3}{2013.2017}\)
\(=\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{2013}-\frac{1}{2017}\right)\)
\(=\frac{3}{4}\left(\frac{1}{5}-\frac{1}{2017}\right)\)
\(=\frac{1509}{10085}\)
TD
0
\(S=\frac{1}{4}\times\left(\frac{4}{5\times9}+\frac{4}{9\times13}+\frac{4}{13\times17}+...+\frac{4}{41\times45}\right)\)
\(S=\frac{1}{4}\times\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)\)
\(S=\frac{1}{4}\times\left(\frac{1}{5}-\frac{1}{45}\right)\)
\(S=\frac{1}{4}\times\frac{8}{45}\)
\(S=\frac{1\times2}{1\times45}\)
\(S=\frac{2}{45}\)
Vậy \(S=\frac{2}{45}\)
Tk nha bn !!
\(\frac{8}{135}\) nhé