DH
2
K
Khách
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NY
3
3 tháng 1 2019
B=1+1/5+1/52+...+1/52018
=>5B=5+1+1/5+...+1/52017
=>5B-B=5-1/52018
=>4B=5-1/52018
=>B=(5-1/52018)/4
3 tháng 1 2019
\(B=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2018}}\)
\(\Rightarrow5B=5\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2018}}\right)\)
\(\Rightarrow5B=5+1+\frac{1}{5}+...+\frac{1}{5^{2017}}\)
\(\Rightarrow5B-B=\left(5+1+\frac{1}{5}+...+\frac{1}{5^{2017}}\right)-\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2018}}\right)\)
\(\Rightarrow4B=5-\frac{1}{5^{2018}}\)
\(\Rightarrow B=\frac{5-\frac{1}{5^{2018}}}{4}\)
Vậy \(B=\frac{5-\frac{1}{5^{2018}}}{4}\)
S
1
27 tháng 9 2021
\(\left(9^2\cdot\dfrac{3^3}{3^7}\right)\cdot2018\)
\(=\dfrac{3^4}{3^4}\cdot2018\)
=2018
14 tháng 10 2023
a) \(12\cdot\left(-\dfrac{2}{3}\right)^2+\dfrac{4}{3}\)
\(=12\cdot\dfrac{4}{9}+\dfrac{4}{3}\)
\(=\dfrac{12\cdot4}{9}+\dfrac{4}{3}\)
\(=\dfrac{16}{3}+\dfrac{4}{3}\)
\(=\dfrac{16+4}{3}\)
\(=\dfrac{20}{3}\)
b) \(\left(\dfrac{3}{2}\right)^2-\left[0,5:2-\sqrt{81}\cdot\left(-\dfrac{1}{2}\right)^2\right]\)
\(=\dfrac{9}{4}-\left(\dfrac{1}{2}:2-9\cdot\dfrac{1}{4}\right)\)
\(=\dfrac{9}{4}-\left(\dfrac{1}{4}-9\cdot\dfrac{1}{4}\right)\)
\(=\dfrac{9}{4}-\dfrac{1}{4}\cdot\left(1-9\right)\)
\(=\dfrac{9}{4}+\dfrac{8}{4}\)
\(=\dfrac{17}{4}\)
c) \(\left(-\dfrac{3}{4}+\dfrac{2}{3}\right):\dfrac{5}{11}+\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\)
\(=-\dfrac{1}{12}:\dfrac{5}{11}+\dfrac{1}{12}\)
\(=\dfrac{1}{12}\cdot-\dfrac{11}{5}+\dfrac{1}{12}\)
\(=\dfrac{1}{12}\cdot\left(-\dfrac{11}{5}+1\right)\)
\(=\dfrac{1}{12}\cdot-\dfrac{6}{5}\)
\(=-\dfrac{1}{10}\)
d) \(\dfrac{\left(-1\right)^3}{15}+\left(-\dfrac{2}{3}\right)^2:2\dfrac{2}{3}-\left|-\dfrac{5}{6}\right|\)
\(=-\dfrac{1}{15}+\dfrac{4}{9}:\left(2+\dfrac{2}{3}\right)-\dfrac{5}{6}\)
\(=-\dfrac{1}{15}+\dfrac{4}{9}:\dfrac{8}{3}-\dfrac{5}{6}\)
\(=-\dfrac{9}{10}+\dfrac{1}{6}\)
\(=-\dfrac{11}{15}\)
e) \(\dfrac{3^7\cdot8^6}{6^6\cdot\left(-2\right)^{12}}\)
\(=\dfrac{3^7\cdot\left(2^3\right)^6}{2^6\cdot3^6\cdot2^{12}}\)
\(=\dfrac{3^7\cdot2^{18}}{2^{6+12}\cdot3^6}\)
\(=\dfrac{2^{18}\cdot3^7}{2^{18}\cdot3^6}\)
\(=3^{7-6}\)
\(=3\)
H
14 tháng 10 2023
\(a,12\cdot\left(-\dfrac{2}{3}\right)^2+\dfrac{4}{3}\\ =12\cdot\dfrac{4}{9}+\dfrac{4}{3}\\ =\dfrac{16}{3}+\dfrac{4}{3}\\ =\dfrac{20}{3}\\ b,\left(\dfrac{3}{2}\right)^2-\left[0,5:2-\sqrt{81}.\left(-\dfrac{1}{2}\right)^2\right]\\ =\dfrac{9}{4}-\left(\dfrac{1}{2}\cdot\dfrac{1}{2}-9\cdot\dfrac{1}{4}\right)\\ =\dfrac{9}{4}-\left(\dfrac{1}{4}-\dfrac{9}{4}\right)\\ =\dfrac{9}{4}-\left(-\dfrac{8}{4}\right)\\ =\dfrac{17}{4}\)
\(c,\left(-\dfrac{3}{4}+\dfrac{2}{3}\right):\dfrac{5}{11}+\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\\ =\left(-\dfrac{9}{12}+\dfrac{8}{12}\right)\cdot\dfrac{11}{5}+\left(-\dfrac{3}{12}+\dfrac{4}{12}\right)\\ =-\dfrac{1}{12}\cdot\dfrac{11}{5}+\dfrac{1}{12}\\ =-\dfrac{11}{60}+\dfrac{1}{12}\\ =-\dfrac{1}{10}\)
\(d,\dfrac{-1^3}{15}+\left(-\dfrac{2}{3}\right)^2:2\dfrac{2}{3}-\left(-\dfrac{5}{6}\right)\\ =-\dfrac{1}{15}+\dfrac{4}{9}\cdot\dfrac{3}{8}+\dfrac{5}{6}\\ =-\dfrac{1}{15}+\dfrac{1}{6}+\dfrac{5}{6}\\ =\dfrac{1}{10}+\dfrac{5}{6}\\ =\dfrac{14}{15}\)
`e,` Không hiểu đề á c: )
11 tháng 8 2023
a) Lập bảng
| n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | ... |
| 7n | 7 | 9 | 3 | 1 | 7 | 9 | 3 | 1 | ... |
| 9n | 9 | 1 | 9 | 1 | 9 | 1 | 9 | 1 | ... |
Ta có: 2018 : 4 = 504 (dư 2)
Suy ra \(2017^{2018}+2019^{2018}= \overline{...9}+\overline{...1}=\overline{...0}\)
Vậy 20172018 + 20192018 chia hết cho 10
b) Làm tương tự như câu a)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
20 tháng 2 2024
Câu 1:
1; 125 : 52
= 53 : 52
= 51
2; 275 : 813
= (33)5 : (34)3
= 315 : 312
= 33
3; 84.165.32
= (23)4.(24)5.25
= 212.220.25
= 237
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
20 tháng 2 2024
Câu 1
4; 274.8110
= (33)4.(34)10
= 312.340
= 352
17 tháng 12 2015
a)\(\frac{1}{4}.x=-\frac{1}{3}\)
\(x=-\frac{1}{3}:\frac{1}{4}\)
\(x=-\frac{4}{3}\)
b)\(-\frac{3}{7}+x=\frac{5}{8}\)
\(\text{ }x=\frac{5}{8}-\left(-\frac{3}{7}\right)\)
\(x=\frac{59}{56}\)
c)\(\frac{16}{2^x}=2\)
\(2^x=\frac{16}{2}\)
\(2^x=8\)
\(\Rightarrow2^x=2^3\)
vậy x=3
\(\dfrac{1}{-7^{2018}}:\dfrac{1}{14^{2018}}=\dfrac{-1}{7^{2018}}\cdot14^{2018}\)
\(=\dfrac{-7^{2018}\cdot2^{2018}}{7^{2018}}=-2^{2018}\)
\(\dfrac{1}{\left(-7\right)^{2018}}:\dfrac{1}{14^{2018}}=\dfrac{1}{7^{2018}}\times14^{2018}\)
\(=\left(\dfrac{14}{7}\right)^{2018}=2^{2018}\)