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\(\dfrac{1}{5}+\dfrac{1}{5^2}+.................+\dfrac{1}{5^{10}}\)
K
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Những câu hỏi liên quan
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
6 tháng 8 2023
\(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{3+2}{6}-\dfrac{1}{5}=\dfrac{5}{6}-\dfrac{1}{5}=\dfrac{5\times5-6}{30}=\dfrac{19}{30}\\ \dfrac{1}{5}:4+\dfrac{3}{4}=\dfrac{1}{5}\times4+\dfrac{3}{4}=\dfrac{4}{5}+\dfrac{3}{4}=\dfrac{4\times4+3\times5}{20}=\dfrac{31}{20}\)
\(\dfrac{4}{3}\times\dfrac{9}{5}-\dfrac{3}{10}=\dfrac{36}{15}-\dfrac{3}{10}=\dfrac{12}{5}-\dfrac{3}{10}=\dfrac{12\times2-3}{10}=\dfrac{21}{10}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
6 tháng 8 2023
\(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\)
= \(\dfrac{1\times15}{2\times15}\) + \(\dfrac{1\times10}{3\times10}\) - \(\dfrac{1\times6}{5\times6}\)
= \(\dfrac{15}{30}\) + \(\dfrac{10}{30}\) - \(\dfrac{6}{30}\)
= \(\dfrac{19}{30}\)
\(\dfrac{1}{5}\) : 4 + \(\dfrac{3}{4}\)
= \(\dfrac{1}{5}\) \(\times\) \(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)
= \(\dfrac{1}{20}\) + \(\dfrac{3\times5}{4\times5}\)
= \(\dfrac{1}{20}+\dfrac{15}{20}\)
= \(\dfrac{16}{20}\)
= \(\dfrac{4}{5}\)
\(\dfrac{4}{3}\) \(\times\) \(\dfrac{9}{5}\) - \(\dfrac{3}{10}\)
= \(\dfrac{12}{5}\) - \(\dfrac{3}{10}\)
= \(\dfrac{24}{10}\) - \(\dfrac{3}{10}\)
= \(\dfrac{21}{10}\)
K
7 tháng 4 2022
\(a,\dfrac{13}{14}\cdot\dfrac{-7}{8}+\dfrac{-3}{2}\)
\(=-\dfrac{13}{16}+\dfrac{-3}{2}\)
\(=-\dfrac{13}{16}+\dfrac{-24}{16}\)
\(=-\dfrac{37}{16}\)
\(b,\dfrac{5}{17}+\dfrac{-15}{34}\cdot\dfrac{2}{5}\)
\(=\dfrac{5}{17}+\dfrac{-3}{17}\)
\(=\dfrac{2}{17}\)
\(c,\dfrac{1}{5}:\dfrac{1}{10}-\dfrac{1}{3}\cdot\left(\dfrac{6}{5}-\dfrac{2}{4}\right)\)
\(=2-\dfrac{1}{3}\cdot\dfrac{7}{10}\)
\(=2-\dfrac{7}{30}\)
\(=\dfrac{53}{30}\)
\(d,\dfrac{-3}{4}:\left(\dfrac{12}{-5}-\dfrac{-7}{10}\right)\)
\(=\dfrac{-3}{4}:\dfrac{-17}{10}\)
\(=\dfrac{15}{34}\)
18 tháng 9 2023
\(3\dfrac{1}{5}\times\dfrac{10}{11}+1\dfrac{2}{11}\\ =\dfrac{16}{5}\times\dfrac{10}{11}+\dfrac{13}{11}\\ =\dfrac{16\times10}{5\times11}+\dfrac{13}{11}\\ =\dfrac{160}{55}+\dfrac{13}{11}\\ =\dfrac{32}{11}+\dfrac{13}{11}\\ =\dfrac{32+13}{11}\\ =\dfrac{45}{11}\)
\(5\dfrac{1}{3}:1\dfrac{2}{3}-1\dfrac{1}{5}\\ =\dfrac{16}{3}:\dfrac{5}{3}-\dfrac{6}{5}\\ =\dfrac{16}{3}\times\dfrac{3}{5}-\dfrac{6}{5}\\ =\dfrac{16\times3}{3\times5}-\dfrac{6}{5}\\ =\dfrac{48}{15}-\dfrac{6}{5}\\ =\dfrac{16}{5}-\dfrac{6}{5}\\ =\dfrac{16-6}{5}\\ =\dfrac{10}{5}\\ =2\)
18 tháng 9 2023
\(3\dfrac{1}{5}\times\dfrac{10}{11}+1\dfrac{2}{11}\\ =\dfrac{3\times5+1}{5}\times\dfrac{10}{11}+\dfrac{1\times11+2}{11}\\ =\dfrac{16}{5}\times\dfrac{10}{11}+\dfrac{13}{11}\\ =\dfrac{160}{55}+\dfrac{13}{11}\\ =\dfrac{32}{11}+\dfrac{13}{11}\\ =\dfrac{45}{11}\)
\(5\dfrac{1}{3}:1\dfrac{2}{3}-1\dfrac{1}{5}\\ =\dfrac{5\times3+1}{3}:\dfrac{1\times3+2}{3}-\dfrac{1\times5+1}{5}\\ =\dfrac{16}{3}:\dfrac{5}{3}-\dfrac{6}{5}\\ =\dfrac{16}{3}\times\dfrac{3}{5}-\dfrac{6}{5}\\ =\dfrac{16}{5}-\dfrac{6}{5}\\ =\dfrac{10}{5}=2\)
10 tháng 3 2022
\(2\dfrac{1}{3}.3=\dfrac{7}{3}.3=7.\\ \left(\dfrac{2}{5}-\dfrac{3}{4}\right)-\dfrac{2}{5}=\dfrac{2}{5}-\dfrac{3}{4}-\dfrac{2}{5}=-\dfrac{3}{4}.\\ \dfrac{-10}{11}.\dfrac{4}{7}+\dfrac{-10}{11}.\dfrac{3}{7}+1\dfrac{10}{11}.\\ =\dfrac{-10}{11}\left(\dfrac{4}{7}+\dfrac{3}{7}-1\right).\\ =\dfrac{-10}{11}.\left(1-1\right)=0.\)
HN
10 tháng 3 2022
1) 2\(\dfrac{1}{3}\).3=\(\dfrac{7}{3}\).3=7.
2) (2/5 -3/4) -2/5 = 2/5 -3/4 -2/5 = -3/4.
3) \(\dfrac{-10}{11}.\dfrac{4}{7}+\dfrac{-10}{11}.\dfrac{3}{7}+1\dfrac{10}{11}=\dfrac{1}{11}\left(-\dfrac{40}{7}-\dfrac{30}{7}+21\right)=\dfrac{1}{11}.\left(-10+21\right)=1\).
H9
HT.Phong (9A5)
CTVHS
24 tháng 8 2023
a) \(3\times\dfrac{4}{11}=\dfrac{3\times4}{11}=\dfrac{12}{11}\)
b) \(1\times\dfrac{5}{4}=\dfrac{1\times5}{4}=\dfrac{5}{4}\)
c) \(0\times\dfrac{2}{5}=\dfrac{0\times2}{5}=\dfrac{0}{5}=0\)
24 tháng 8 2023
a: \(=\dfrac{3\cdot4}{11}=\dfrac{12}{11}\)
b: \(=\dfrac{1\cdot5}{4}=\dfrac{5}{4}\)
c: \(=\dfrac{0\cdot2}{5}=0\)
TM
16 tháng 3 2021
câu b bài 2:
\(\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\)
\(=\dfrac{1}{5}\)
câu a bài 2:
\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{10\cdot11\cdot12}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}-...-\dfrac{1}{12}\)
\(=1-\dfrac{1}{12}=\dfrac{11}{12}\)
12 tháng 3 2023
Sửa đề : \(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{1\dfrac{1}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\)
\(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{1\dfrac{1}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\\ =\left(\dfrac{2\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{7}{11}\right)}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\\ =\left(\dfrac{2}{7}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}\right)}\right):\dfrac{2021}{2020}\\ =\left(\dfrac{2}{7}-\dfrac{2}{7}\right):\dfrac{2021}{2022}=0\)
Đặt \(A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{10}}\)
\(\Rightarrow5A=1+\dfrac{1}{5}+...+\dfrac{1}{5^9}\)
\(\Rightarrow5A-A=\left(1+\dfrac{1}{5}+...+\dfrac{1}{5^9}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{10}}\right)\)
\(\Rightarrow4A=1-\dfrac{1}{5^{10}}\)
\(\Rightarrow A=\dfrac{1-\dfrac{1}{5^{10}}}{4}\)
Vậy \(\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{10}}=\dfrac{1-\dfrac{1}{5^{10}}}{4}\)
\(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+......+\dfrac{1}{5^{10}}\)
= \(\dfrac{1}{5}\left(1+\dfrac{1}{5}+\dfrac{1}{5^2}+....+\dfrac{1}{5^9}\right)\)
= \(\left[\left(1+\dfrac{1}{5}+\dfrac{1}{5^2}+....+\dfrac{1}{5^9}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+....+\dfrac{1}{5^{10}}\right)\right]\) : 4
= \(\left(1-\dfrac{1}{10}\right):4\)