Tính:(1/7)^5.7^6
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
12 tháng 10 2019
A=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)
=\(\frac{2}{1}-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{49}-\frac{2}{51}\)
= \(2.(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51})\)
=2.\((1-\frac{1}{51})\)
=\(2.\frac{50}{51}\)
=\(\frac{100}{51}\)
VD
6 tháng 10 2023
-3^7.2^8/2^.3^7
=-3.2
=-6
5^3.3^5/5^3(0,5+2,5)
=5^3.3^5/5^3.3\
3^4
=81
5.7^4+7^3.25/7^5.125-7^3.50
=5.7^3(7+5
VD
6 tháng 10 2023
5.7^4+7^3.25/7^5.125-7^3.50
=5.7^4+7^3.5^2/7^5.5^3-7^3.11.5
=5.7^3(1.7+1.5)/7^3.5(7^2.25-11)
12/1250
HN
27 tháng 3 2018
Câu 1:
\(A=\frac{2^5.7+2^5}{2^5.3-2^5}\)= \(\frac{2^5.8}{2^5.2}\)= 4
Vậy A = 4
Câu 2:
\(B=2^3.5^3-3.\left\{400-\left[673-2^3.\left(7^8:7^6+7^0\right)\right]\right\}\)
\(B=8.125-3.\left\{400-\left[673-8.\left(7^2+1\right)\right]\right\}\)
\(B=1000-3.\left\{400-\left[673-8.\left(49+1\right)\right]\right\}\)
\(B=1000-3.\left\{400-\left[673-8.50\right]\right\}\)
\(B=1000-3.\left\{400-\left[673-400\right]\right\}\)
\(B=1000-3.\left\{400-273\right\}\)
\(B=1000-3.127\)
\(B=1000-381\)
\(B=619\)
Vậy B = 619
U
27 tháng 7 2018
\(a,\frac{6}{5\cdot7}+\frac{6}{7\cdot9}+\frac{6}{9\cdot11}+...+\frac{6}{103\cdot105}\)
\(=\frac{6}{2}\left(\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{103\cdot105}\right)\)
\(=\frac{6}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{103}-\frac{1}{105}\right)\)
\(=\frac{6}{2}\left(\frac{1}{5}-\frac{1}{105}\right)\)
\(=\frac{6}{2}\cdot\frac{20}{105}\)
\(=\frac{60}{105}\)
\(b,\left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{8}\right)\left(1-\frac{1}{10}\right)\)
\(=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{7}{8}\cdot\frac{9}{10}\)
\(=\frac{189}{640}\)
\(c,\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right)\left(1-\frac{1}{9}\right)\)
\(=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\frac{8}{9}\)
\(=\frac{384}{945}\)
IM
29 tháng 7 2016
a)
Số số hạng là
\(\left(40-1\right):3+1=14\)
Tổng là
\(S=\frac{\left(40+1\right).14}{2}=287\)
9 tháng 5 2016
B=\(\frac{6}{1.3}+\frac{6}{3.5}+\frac{6}{5.7}+......+\frac{6}{99.101}\)
=\(6.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.......+\frac{1}{99.101}\right)\)
=\(6\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+......+\frac{1}{99}-\frac{1}{101}\right)\)
=\(6.\left(1-\frac{1}{101}\right)\)
=\(6.\frac{100}{101}\)
=\(\frac{600}{101}\)
XO
26 tháng 8 2019
Tính :
a) \(M=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
b) \(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=7.\frac{3}{35}\)
\(=\frac{3}{5}\)
c) \(B=\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)
\(=\frac{1}{2}.\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)\)
\(=\frac{1}{2}.\frac{2}{75}\)
\(=\frac{1}{75}\)
SM
1
7 tháng 8 2019
b) \(\frac{-5\cdot7^5+7^4}{7^6\cdot10-2\cdot7^5}\)
\(=\frac{-35\cdot7^4+7^4}{7^5\cdot70-2\cdot7^5}\)
\(=\frac{7^4\left(-35+1\right)}{7^5\left(70-2\right)}\)
\(=\frac{7^4\cdot\left(-34\right)}{7^5\cdot68}\)
\(=\frac{-1}{14}\)
Chắc sai =))
L
2
TN
18 tháng 4 2016
đặt A=6/5.7+6/7.9+6/9.11+......+6/33.35
\(\Leftrightarrow A=3\left(\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{33\cdot35}\right)\)
\(\Rightarrow A=3\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{33}-\frac{1}{35}\right)\)
\(\Rightarrow A=3\left(\frac{1}{5}-\frac{1}{35}\right)\)
\(\Rightarrow A=3\cdot\frac{6}{35}\)
\(\Rightarrow A=\frac{18}{35}\)
BI
18 tháng 4 2016
Gọi A là biểu thức \(\frac{6}{5\cdot7}+\frac{6}{7\cdot9}+...+\frac{6}{33\cdot35}\)
Ta có: \(\frac{6}{5\cdot7}+\frac{6}{7\cdot9}+...+\frac{6}{33\cdot35}\)
\(\Rightarrow A=3\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{33}-\frac{1}{35}\right)\)
\(\Rightarrow A=3\left(\frac{1}{5}-\frac{1}{35}\right)=3\cdot\frac{6}{35}=\frac{18}{35}\)
( \(\dfrac{1}{7}\))5.76 = \(\dfrac{7^6}{7^5}\) = 76-5 = 7