Tính các tổng sau:
a) \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}.\)
b) \(-\frac{4}{5}+\frac{4}{5^2}-\frac{4}{5^3}+...+\frac{4}{5^{200}}.\)
c)\(\frac{-1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\)
Tính các tổng sau
\(A=7-7^4+7^7-...+7^{301}\)
\(B=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
\(C=\frac{4}{5}+\frac{4}{5^2}-\frac{4}{5^3}+...+\frac{4}{5^{200}}\)
AI TÍNH ĐƯỢC MÌNH CHO 10 TICK NHA
Tính
A = \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
B = \(\frac{4}{5}+\frac{4}{5^2}-\frac{4}{5^3}+...+\frac{4}{5^{200}}\)
A=\(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
\(\Rightarrow7A=(1+\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{99}})-\left(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+....+\frac{1}{7^{100}}\right)\)
\(\Rightarrow6A=\left(1-\frac{1}{7^{99}}\right)\)
\(\Rightarrow A=\left(1-\frac{1}{7^{99}}\right):6\)
Câu b tương tự nha
a) \(A=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...........+\frac{1}{7^{100}}\)
\(\Rightarrow7A=1+\frac{1}{7}+\frac{1}{7^2}+.........+\frac{1}{7^{99}}\)
\(\Rightarrow7A-A=6A=1-\frac{1}{7^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{7^{100}}}{6}\)
\(A=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
\(\frac{A}{7}=\frac{1}{7^2}+\frac{1}{7^3}+\frac{1}{7^4}+...+\frac{1}{7^{101}}\)
\(A-\frac{A}{7}=\left(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\right)-\left(\frac{1}{7^2}+\frac{1}{7^3}+\frac{1}{7^4}+...+\frac{1}{7^{101}}\right)\)
\(\frac{6}{7}A=\frac{1}{7}-\frac{1}{7^{101}}\)
\(A=\left(\frac{1}{7}-\frac{1}{7^{101}}\right).\frac{7}{6}\)
\(A=\frac{1}{6}-\frac{1}{6.7^{100}}\)
\(B=\frac{4}{5}+\frac{4}{5^2}-\frac{4}{5^3}+...+\frac{4}{5^{200}}\)
\(=4.\left(\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^3}+...+\frac{1}{5^{200}}\right)\)
Gọi \(C=\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^3}+...+\frac{1}{5^{200}}\)
\(\frac{C}{5}=\frac{1}{5^2}+\frac{1}{5^3}-\frac{1}{5^4}+...+\frac{1}{5^{201}}\)
\(C-\frac{5}{C}=\left(\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^3}+...+\frac{1}{5^{200}}\right)-\left(\frac{1}{5^2}+\frac{1}{5^3}-\frac{1}{5^4}+...+\frac{1}{5^{201}}\right)\)
\(\frac{4}{5}C=\frac{1}{5}-\frac{1}{5^{201}}\)
\(C=\left(\frac{1}{5}-\frac{1}{5^{201}}\right).\frac{5}{4}\)
\(=\frac{1}{4}-\frac{1}{4.5^{200}}\)
Thay vào B ta có
\(B=4.\left(\frac{1}{4}-\frac{1}{4.5^{200}}\right)\)
=\(=1-\frac{1}{5^{200}}\)
thực hiện phép tính :
a, \(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{\frac{7}{6}-58+5+0,7}\)
b, \(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
c, \(\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)
Mong các bạn giúp đỡ nhé
Tinh cac tong sau
\(H=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
\(I=\frac{4}{5}+\frac{4}{5^2}+\frac{4}{5^3}+...+\frac{4}{5^{200}}\)
thực hiên các phép tính tính :
a) \(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}}\)
b) \(\frac{\left(1+2+3+...+100\right).\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(6,3.12-21.3,6\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
Bài 1:Thực hiện các phép tính
a)A=\(1-\frac{1}{1+\frac{2}{1-\frac{3}{1-4}}}\)
b)B=\(-\frac{1}{10}-\frac{1}{100}-\frac{1}{1000}-\frac{1}{10000}-\frac{1}{100000}-\frac{1}{1000000}\)
Bài 2:Thực hiện các phép tính sau 1 cách hợp lý
a)A=\(\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)
b)B=\(\frac{\frac{2}{39}-\frac{1}{15}-\frac{2}{153}}{\frac{1}{34}+\frac{3}{20}-\frac{3}{26}}:\frac{1+\frac{2}{71}-\frac{5}{121}}{\frac{65}{121}-\frac{26}{71}-13}\)
c)C=\(\left(\frac{112}{13.20}+\frac{112}{20.27}+\frac{112}{27.34}+...+\frac{112}{62.69}\right):\left(-\frac{5}{9.13}-\frac{7}{9.25}-\frac{13}{19.25}-\frac{31}{19.69}\right)\)
d)D=\(\frac{2.2016}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2016}}\)
e)E=\(-1-\frac{1}{3}-\frac{1}{6}-\frac{1}{10}-\frac{1}{15}-...-\frac{1}{1225}\)
tính nhanh:
A=\(\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)
giải chi tiết nhé các bạn
A=\(\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)
\(=\frac{3\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)
\(=\frac{3}{5}+\frac{1}{-7}=\frac{3}{5}-\frac{1}{7}\)
\(=\frac{21}{35}-\frac{5}{35}=\frac{16}{35}\)
1. Quy đồng mẫu các phân số sau:
a) \(\frac{5}{{12}}\) và \(\frac{7}{{15}}\); b) \(\frac{2}{7};\,\,\frac{4}{9}\) và \(\frac{7}{{12}}\).
2. Thực hiện các phép tính sau:
a) \(\frac{3}{8} + \frac{5}{{24}};\) b) \(\frac{7}{{16}} - \frac{5}{{12}}.\)
1. a) Ta có BCNN(12, 15) = 60 nên ta lấy mẫu chung của hai phân số là 60.
Thừa số phụ:
60:12 =5; 60:15=4
Ta được:
\(\frac{5}{{12}} = \frac{{5.5}}{{12.5}} = \frac{{25}}{{60}}\)
\(\frac{7}{{15}} = \frac{{7.4}}{{15.4}} = \frac{{28}}{{60}}\)
b) Ta có BCNN(7, 9, 12) = 252 nên ta lấy mẫu chung của ba phân số là 252.
Thừa số phụ:
252:7 = 36; 252:9 = 28; 252:12 = 21
Ta được:
\(\frac{2}{7} = \frac{{2.36}}{{7.36}} = \frac{{72}}{{252}}\)
\(\frac{4}{9} = \frac{{4.28}}{{9.28}} = \frac{{112}}{{252}}\)
\(\frac{7}{{12}} = \frac{{7.21}}{{12.21}} = \frac{{147}}{{252}}\)
2. a) Ta có BCNN(8, 24) = 24 nên:
\(\frac{3}{8} + \frac{5}{{24}} = \frac{{3.3}}{{8.3}} + \frac{5}{{24}} = \frac{9}{{24}} + \frac{5}{{24}} = \frac{{14}}{{24}} = \frac{7}{{12}}\)
b) Ta có BCNN(12, 16) = 48 nên:
\(\frac{7}{{16}} - \frac{5}{{12}} = \frac{{7.3}}{{16.3}} - \frac{{5.4}}{{12.4}} = \frac{{21}}{{48}} - \frac{{20}}{{48}} = \frac{1}{{48}}\).
\(\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(\frac{1}{2}+\frac{3}{4}-\left(\frac{3}{4}-\frac{4}{5}\right)\)
\(6\frac{5}{12}:2\frac{3}{4}+11\frac{1}{4}\left(\frac{1}{3}-\frac{1}{5}\right)\)
\(\left(\frac{7}{8}-\frac{3}{4}\right)1\frac{1}{3}-\frac{2}{7}\left(3,5\right)^2\)
\(\left(\frac{3}{5}+0,415-\frac{3}{200}\right)2\frac{2}{3}.0,25\)
Thực hiện các phép tính
cảm ơn
\(\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(=\left(\frac{9}{24}+-\frac{18}{24}+\frac{14}{24}\right):\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}.\frac{6}{5}+\frac{1}{2}\)
\(=\frac{1}{4}+\frac{1}{2}\)
\(=\frac{1}{4}+\frac{2}{4}\)
\(=\frac{3}{4}\)
\(\frac{1}{2}+\frac{3}{4}-\left(\frac{3}{4}-\frac{4}{5}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\left(\frac{15}{20}-\frac{16}{20}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\frac{-1}{20}\)
\(=\frac{10}{20}+\frac{15}{20}-\frac{-1}{20}\)
\(=\frac{25}{20}-\frac{-1}{20}\)
\(=\frac{26}{20}\)
\(=\frac{13}{10}\)
1/2+3/4-(3/4-4/5)
1/2+3/4+3/4+4/5
1/2+6/4+4/5
10/20+30/20+16/20
56/20=14/5