Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\dfrac{7}{10}+\dfrac{7}{10^2}+\dfrac{7}{10^3}+...+\dfrac{7}{10^{2011}}\)
\(\Rightarrow10A=7+\dfrac{7}{10}+\dfrac{7}{10^2}+...+\dfrac{7}{10^{2010}}\)
\(\Rightarrow10A-A=7+\dfrac{7}{10}+\dfrac{7}{10^2}+...+\dfrac{7}{10^{2010}}-\left(\dfrac{7}{10}+\dfrac{7}{10^2}+\dfrac{7}{10^3}+...+\dfrac{7}{10^{2011}}\right)\)
\(\Rightarrow9A=7-\dfrac{7}{10^{2011}}\)
\(\Rightarrow A=\dfrac{7}{9}.\left(1-\dfrac{1}{10^{2011}}\right)\)
\(\frac{3}{7.10}+\frac{3}{10.13}+....+\frac{3}{100.103}\)
\(=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+....+\frac{1}{100}-\frac{1}{103}\)
\(=\frac{1}{7}-\frac{1}{103}\)
\(=\frac{96}{721}\)
\(\frac{2}{7.10}+\frac{2}{10.13}+...+\frac{2}{100.103}\)
\(=\frac{2}{3}\left(\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(=\frac{2}{3}\left(\frac{1}{7}-\frac{1}{103}\right)\)
\(=\frac{2}{3}.\frac{96}{721}\)
\(=\frac{64}{721}\)
A, 910 -4/910- 5
= (9-4/9)10- 5
= 77/910 - 5
910 - 2/910 - 3
=( 9-2/9 )10 - 3
= 79/910 -3
vì 77/9
a) Ta có: \(1-\frac{9^{10}-4}{9^{10}-5}=\frac{-1}{9^{10}-5}\)
\(1-\frac{9^{10}-2}{9^{10}-3}=\frac{-1}{9^{10}-3}\)
Vì \(\frac{-1}{9^{10}-5}< \frac{-1}{9^{10}-3}\Rightarrow1-\frac{9^{10}-4}{9^{10}-5}< 1-\frac{9^{10}-2}{9^{10}-3}\)
\(\Rightarrow\frac{9^{10}-4}{9^{10}-5}>\frac{9^{10}-2}{9^{10}-3}\).
b) Ta có: \(1-\frac{2.7^{10}-1}{7^{10}}=\frac{7^{10}+1}{7^{10}}\)
\(1-\frac{2.7^{10}+1}{7^{10}+1}=\frac{7^{10}}{7^{10}+1}\)
Vì \(\frac{7^{10}+1}{7^{10}}>\frac{7^{10}}{7^{10}+1}\Rightarrow1-\frac{2.7^{10}-1}{7^{10}}>1-\frac{2.7^{10}+1}{7^{10}+1}\)
\(\Rightarrow\frac{2.7^{10}-1}{7^{10}}< \frac{2.7^{10}+1}{7^{10}+1}\)
I don't now
or no I don't
..................
sorry
`@` `\text {Ans}`
`\downarrow`
`a)`
\(\dfrac{7}{5}\cdot\dfrac{8}{19}+\dfrac{7}{5}\cdot\dfrac{12}{19}-\dfrac{7}{5}\cdot\dfrac{1}{19}\)
`=`\(\dfrac{7}{5}\cdot\left(\dfrac{8}{19}+\dfrac{12}{19}-\dfrac{1}{19}\right)\)
`=`\(\dfrac{7}{5}\cdot\dfrac{19}{19}=\dfrac{7}{5}\cdot1=\dfrac{7}{5}\)
`b)`
\(-\dfrac{3}{5}\cdot\dfrac{5}{7}+\left(-\dfrac{3}{5}\right)\cdot\dfrac{3}{7}+\left(-\dfrac{3}{5}\right)\cdot\dfrac{6}{7}\)
`=`\(-\dfrac{3}{5}\cdot\left(\dfrac{5}{7}+\dfrac{3}{7}+\dfrac{6}{7}\right)\)
`=`\(-\dfrac{3}{5}\cdot\dfrac{14}{7}\)
`=`\(-\dfrac{3}{5}\cdot2=-\dfrac{6}{5}\)
`c)`
\(10\dfrac{2}{9}+\left(2\dfrac{2}{5}-7\dfrac{2}{9}\right)\)
`=`\(10\dfrac{2}{9}+2\dfrac{2}{5}-7\dfrac{2}{9}\)
`=`\(\left(10\dfrac{2}{9}-7\dfrac{2}{9}\right)+2\dfrac{2}{5}\)
`=`\(3+2\dfrac{2}{5}=\dfrac{27}{5}\)
`d)`
\(6\dfrac{3}{10}-\left(3\dfrac{4}{7}+2\dfrac{3}{10}\right)\)
`=`\(6\dfrac{3}{10}-3\dfrac{4}{7}-2\dfrac{3}{10}\)
`=`\(\left(6\dfrac{3}{10}-2\dfrac{3}{10}\right)-3\dfrac{4}{7}\)
`=`\(4-3\dfrac{4}{7}=\dfrac{3}{7}\)
a) \(\dfrac{7}{5}.\dfrac{8}{19}+\dfrac{7}{5}.\dfrac{12}{19}-\dfrac{7}{5}.\dfrac{1}{19}\)
\(=\dfrac{7}{5}.\left(\dfrac{8}{19}+\dfrac{12}{19}-\dfrac{1}{19}\right)\)
\(=\dfrac{7}{5}.1\)
\(=\dfrac{7}{5}\)
b) \(\dfrac{-3}{5}.\dfrac{5}{7}+\dfrac{-3}{5}.\dfrac{3}{7}+\dfrac{-3}{5}.\dfrac{6}{7}\)
\(=\dfrac{-3}{5}.\left(\dfrac{5}{7}+\dfrac{3}{7}+\dfrac{6}{7}\right)\)
\(=\dfrac{-3}{5}.2\)
\(=\dfrac{-6}{5}\)
c) \(10\dfrac{2}{9}+\left(2\dfrac{2}{5}-7\dfrac{2}{9}\right)\)
\(=\dfrac{92}{9}+\dfrac{12}{5}-\dfrac{65}{9}\)
\(=\dfrac{92}{9}-\dfrac{65}{9}+\dfrac{12}{5}\)
\(=3+\dfrac{12}{5}\)
\(=\dfrac{15}{5}+\dfrac{12}{5}\)
\(=\dfrac{27}{5}\)
d) \(6\dfrac{3}{10}-\left(3\dfrac{4}{7}+2\dfrac{3}{10}\right)\)
\(=\dfrac{63}{10}-\dfrac{25}{7}-\dfrac{23}{10}\)
\(=\dfrac{63}{10}-\dfrac{23}{10}-\dfrac{25}{7}\)
\(=4-\dfrac{25}{7}\)
\(=\dfrac{28}{7}-\dfrac{25}{7}\)
\(=\dfrac{3}{7}\)
Chúc bạn học tốt
