\(\dfrac{\left(-7\right)^n}{\left(-7\right)^{n-1}}\left(n>hoặc=1\right)\)
K
Khách
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19 tháng 5 2022
a: \(=\left(-\dfrac{5}{7}\right)^{n-n}=\left(-\dfrac{5}{7}\right)^0=1\)
b: \(=\left(-\dfrac{1}{2}\right)^{2n-n}=\left(-\dfrac{1}{2}\right)^n\)
Y
23 tháng 7 2018
a) \(\dfrac{\left(-\dfrac{5}{7}\right)^n}{\left(-\dfrac{5}{7}\right)^{n-1}}\)
\(=\dfrac{\left(-\dfrac{5}{7}\right)^n}{\left(-\dfrac{5}{7}\right)^n:\left(-\dfrac{5}{7}\right)}\)
\(=\dfrac{\left(-\dfrac{5}{7}\right)^n}{\left(-\dfrac{5}{7}\right)^n.\left(-\dfrac{7}{5}\right)}\)
\(=\dfrac{1}{\left(-\dfrac{7}{5}\right)}\)
\(=1.\left(-\dfrac{5}{7}\right)\)
\(=-\dfrac{5}{7}\)
b) \(\dfrac{\left(-\dfrac{1}{2}\right)^{2n}}{\left(-\dfrac{1}{2}\right)^n}\)
\(=\dfrac{\left(-\dfrac{1}{2}\right)^n.\left(-\dfrac{1}{2}\right)^n}{\left(-\dfrac{1}{2}\right)^n}\)
\(=\left(-\dfrac{1}{2}\right)^n\)
29 tháng 8 2017
A = \(\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)
Xét: n4 + 4 = (n2+2)2 - 4n2 = (n2-2n+2)(n2+2n+2) = [(n-1)2+1][(x+1)2+1] nên: A = \(\dfrac{\left(0^2+1\right)\left(2^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)}.\dfrac{\left(4^2+1\right)\left(6^2+1\right)}{\left(6^2+1\right)\left(8^2+1\right)}.....\dfrac{\left(20^2+1\right)\left(22^2+1\right)}{\left(22^2+1\right)\left(24^2+1\right)}=\dfrac{1}{24^2+1}=\dfrac{1}{577}\)
B = \(\left(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{2}{n-2}+\dfrac{1}{n-1}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\right)\)
Đặt C = \(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{n-\left(n-2\right)}{n-2}+\dfrac{n-\left(n-1\right)}{n-1}\)
= \(\dfrac{n}{1}+\dfrac{n}{2}+...+\dfrac{n}{n-2}+\dfrac{n}{n-1}-1-1-...-1\)
= \(n+\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}-\left(n-1\right)\)
= \(\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}+\dfrac{n}{n}\)
= \(n\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\right)\)
Vậy ...
9 tháng 4 2018
\(A=\dfrac{3}{\left(1\cdot2\right)^2}+\dfrac{5}{\left(2\cdot3\right)^2}+\dfrac{7}{\left(3\cdot4\right)^2}+...+\dfrac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
\(A=\dfrac{3}{1\cdot4}+\dfrac{5}{4\cdot9}+\dfrac{7}{9\cdot16}+...+\dfrac{2n+1}{n^2\cdot\left(n^2+2n+1\right)}\)
\(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{n^2}-\dfrac{1}{n^2+2n+1}\)
\(A=1-\dfrac{1}{n^2+2n+1}\)
\(A=\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\)
AH
Akai Haruma
Giáo viên
8 tháng 3 2021
Đề bị lỗi công thức rồi bạn. Bạn cần viết lại để được hỗ trợ tốt hơn.
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 1 2021
Chia cả tử và mẫu cho \(n^5\)
\(=\lim\dfrac{\left(\dfrac{2n-n^3}{n^3}\right)\left(\dfrac{3n^2+1}{n^2}\right)}{\left(\dfrac{2n-1}{n}\right)\left(\dfrac{n^4-7}{n^4}\right)}=\lim\dfrac{\left(\dfrac{2}{n^2}-1\right)\left(3+\dfrac{1}{n^2}\right)}{\left(2-\dfrac{1}{n}\right)\left(1-\dfrac{7}{n^4}\right)}\)
\(=\dfrac{-1.3}{2.1}=-\dfrac{3}{2}\)
HT
17 tháng 1 2021
Dang này thì cứ chọn số hạng có mũ cao nhất trên tử và mẫu là được. Nó là ngắt vô cùng lớn hay bé gì đấy
\(=lim\dfrac{8n^6}{3n^6}=\dfrac{8}{3}\)
24 tháng 5 2017
\(\left(1+\dfrac{7}{9}\right).\left(1+\dfrac{7}{20}\right).\left(1+\dfrac{7}{33}.\right)\left(1+\dfrac{7}{48}\right)...\left(1+\dfrac{7}{180}\right)\)
\(=\dfrac{16}{9}.\dfrac{27}{20}.\dfrac{40}{33}.\dfrac{55}{48}...\dfrac{7}{180}\)
\(=\dfrac{2.8}{1.9}.\dfrac{3.9}{2.10}.\dfrac{4.10}{3.11}.\dfrac{5.11}{4.12}...\dfrac{11.17}{10.18}\)
\(=\dfrac{\left(2.3.4.5...11\right).\left(8.9.10.11...17\right)}{\left(1.2.3.4...10\right).\left(9.10.11.12...18\right)}\)
\(=\dfrac{11.8}{1.18}=\dfrac{88}{18}=\dfrac{44}{9}\)
24 tháng 5 2017
ta có ;
\(\left(1+\dfrac{7}{9}\right)\cdot\left(1+\dfrac{7}{20}\right).\left(1+\dfrac{7}{33}\right)...\left(1+\dfrac{1}{180}\right)\)
=\(\dfrac{16}{9}.\dfrac{27}{20}.\dfrac{40}{33}....\dfrac{187}{180}\)
=\(\dfrac{8.2}{9.1}.\dfrac{9.3}{10.2}.\dfrac{10.4}{3.11}.\dfrac{11.5}{4.12}....\dfrac{17.11}{18.10}\)
=\(\dfrac{8.9.10.11.12.13.14.15.16.17.2.3.4.5.6.7.8.9.10.11}{9.10.11.12.13.14.15.16.17.18.1.2.3.4.5.6.7.8.9.10}\)
=\(\dfrac{8.11}{18}=\dfrac{88}{18}=\dfrac{44}{9}\)
mình ko hiểu bài này-
28 tháng 10 2017
\(\dfrac{\left(-\dfrac{5}{7}\right)^{n+1}}{\left(-\dfrac{5}{7}\right)^n}=\left(-\dfrac{5}{7}\right)^{n+1}:\left(-\dfrac{5}{7}\right)^n\\ =\left(-\dfrac{5}{7}\right)\)
\(\dfrac{\left(-7\right)^n}{\left(-7\right)^{n-1}}=\dfrac{\left(-7\right)^n}{\left(-7\right)^n\cdot\left(-7\right)^{-1}}=\dfrac{1}{\left(-7\right)^{-1}}=-7\)