\(\frac{3}{2}\)+\(\frac{3}{2^2}\)+
K
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21 tháng 4 2017

\(=\frac{3}{200}:\frac{3}{5}+\frac{3}{2}\left(\frac{4}{25}-\frac{2}{5}\right)-\frac{1}{25}.\left(\frac{7}{4}:\frac{7}{5}-\frac{5}{2}\right)\)

\(=\frac{3.5}{200.3}+\frac{3}{2}\left(\frac{4}{25}-\frac{2.5}{25}\right)-\frac{1}{25}\left(\frac{7.5}{4.7}-\frac{5}{2}\right)\)

\(=\frac{1}{40}+\frac{3}{2}\left(\frac{-6}{25}\right)-\frac{1}{25}\left(\frac{5}{4}-\frac{10}{4}\right)\)

\(=\frac{1}{40}-\frac{9}{25}+\frac{1}{20}\)

\(=\frac{1.5}{40.5}-\frac{9.8}{25.8}+\frac{1.10}{20.10}\)

\(=\frac{5-72+10}{200}=\frac{-57}{200}\)

21 tháng 4 2017

Thử máy tính lại ròi kq đúng

30 tháng 1 2017

S = \(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)

\(=3\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)

Đặt A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)

2A = \(2\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)

\(2+1+\frac{1}{2}+....+\frac{1}{2^8}\)

\(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^8}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)

\(A=2-\frac{1}{2^9}\)

\(\Rightarrow S=3\left(2-\frac{1}{2^9}\right)=\frac{3.\left(2^{10}-1\right)}{2^9}\)

25 tháng 4 2017

\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)

\(2S=6+3+\frac{3}{2}+...+\frac{3}{2^8}\)

\(2S-S=\left(6+3+\frac{3}{2}+...+\frac{3}{2^8}\right)-\left(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\)

\(S=6-\frac{3}{2^9}=6-\frac{3}{512}=\frac{3069}{512}\)

11 tháng 4 2017

\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)

\(\Leftrightarrow S=3.\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)

- Đặt  \(D=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)

\(\Leftrightarrow\frac{1}{2}D=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)

\(\Leftrightarrow\frac{1}{2}D-D=\frac{1}{2^{10}}-1\)

\(\Leftrightarrow D=\frac{\frac{1}{2^{10}}-1}{-\frac{1}{2}}\)

Vậy \(3.D=3.\left(\frac{\frac{1}{2^{10}}-1}{-\frac{1}{2}}\right)=3.\frac{1023}{512}=\frac{3069}{512}\)

11 tháng 4 2017

Ta có: \(\frac{1}{2}S=\frac{1}{2}.\left(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\))

=\(\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^{10}}\)

=> \(S-\frac{1}{2}S=\left(3+\frac{3}{2}+...+\frac{3}{2^9}\right)-\left(\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^{10}}\right)\)

=> \(\frac{1}{2}S=3-\frac{3}{2^{10}}\)

=>\(S=\left(3-\frac{3}{2^{10}}\right).2=6-\frac{6}{2^{10}}=6-\frac{3}{2^9}\)

11 tháng 3 2017

Thực hiện phép tính 

a ) \(\frac{2}{5}+\frac{-1}{6}-\frac{3}{4}-\frac{-2}{3}\)

\(\frac{2}{5}+\frac{-1}{6}+\frac{-3}{4}+\frac{2}{3}\)

\(\left(\frac{2}{5}+\frac{-3}{4}\right)+\left(\frac{-1}{6}+\frac{2}{3}\right)\)

\(\left(\frac{8}{20}+\frac{-15}{20}\right)+\left(\frac{-1}{6}+\frac{4}{6}\right)\)

\(\left(\frac{8+\left(-15\right)}{20}\right)+\left(\frac{\left(-1\right)+4}{6}\right)\)

\(\frac{-7}{20}+\frac{1}{2}\)

\(\frac{-7}{20}+\frac{10}{20}=\frac{\left(7\right)+10}{20}=\frac{3}{20}\)

tk mk nha 

đang âm rất  nhiều rồi  , giúp nha !!!!!

2 tháng 1 2017

nhanh

24 tháng 4 2019

\(A=3\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+.....+\frac{3}{55\cdot58}\right)\)

\(A=3\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{55}-\frac{1}{58}\right)\)

\(A=3\left(1-\frac{1}{58}\right)\)

\(A=3-\frac{1}{174}< 3< \frac{10}{3}\)