\(\frac{1}{2^3}\)D= \(\frac{1}{2^4}-\frac{1}{2^7}+\frac{1}{2^{10}}-\frac{1}{2^{13}}+...+\frac{1}{2^{58}}-\frac{1}{2^{61}}\)

D+ \(\frac{1}{2^3}\)D=\(\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^4}+\frac{1}{2^7}-\frac{1}{2^7}-\frac{1}{2^{10}}+\frac{1}{2^{10}}+...-\frac{1}{2^{58}}+\frac{1}{2^{58}}-\frac{1}{2^{61}}\)

\(\frac{9}{8}\)D= \(\frac{1}{2}-\frac{1}{2^{61}}\)=> D= \(\frac{\frac{1}{2}-\frac{1}{2^{61}}}{\frac{9}{8}}\)

\(D=\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-\frac{1}{2^{10}}+...+\frac{1}{2^{55}}-\frac{1}{2^{58}}\)

\(\Rightarrow2^3D=2^2-\frac{1}{2}+\frac{1}{2^4}-\frac{1}{2^7}+....+\frac{1}{2^{52}}-\frac{1}{2^{55}}\)

\(\Rightarrow8D+D=2^2-\frac{1}{2^{58}}\)

\(\Rightarrow D=\frac{2^2-\frac{1}{2^{58}}}{9}\)

nhân 2 vế cho \(\frac{1}{2^3}\)

\(2A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+.....-\frac{1}{2^{99}}\Rightarrow2A+A=3A=\left(1-\frac{1}{2}+\frac{1}{2^2}-....-\frac{1}{2^{99}}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+......-\frac{1}{2^{100}}\right)=1-\frac{1}{2^{100}}=\frac{2^{100}-1}{2^{100}}\Rightarrow A=\frac{2^{100}-1}{3.2^{100}}\)

\(2,4B=2+\frac{1}{2}+\frac{1}{2^3}+.....+\frac{1}{2^{97}}\Rightarrow4B-B=3B=\left(2+\frac{1}{2}+....+\frac{1}{2^{97}}\right)-\left(\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)=2-\frac{1}{2^{99}}=\frac{2^{100}-1}{2^{99}}\Rightarrow B=\frac{2^{100}-1}{3.2^{99}}\)

\(3,C=\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-....-\frac{1}{2^{58}}\Rightarrow8C=4-\frac{1}{2}+\frac{1}{2^4}-.....-\frac{1}{2^{55}}\Rightarrow8C+C=9C=\left(4-\frac{1}{2}+\frac{1}{2^4}-....-\frac{1}{2^{55}}\right)+\left(\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-....-\frac{1}{2^{58}}\right)=4-\frac{1}{2^{58}}=\frac{2^{60}-1}{2^{58}}\Rightarrow C=\frac{2^{60}-1}{9.2^{58}}\)

a) 20,7 + 1,47 : 7 - 0,23 . 5

= 20,7 + 0,21 – 1,15

= 20,91 – 1,15

= 19,76

Ở trên vietjack có đó bn =)

a, 20,7 + 1,47 : 7 - 0,23 . 5

=\(\frac{207}{10}+\frac{147}{100}:7-\frac{23}{100}.5\)

= \(\frac{207}{10}+\frac{21}{100}-\frac{23}{20}\)

= \(\frac{2091}{100}+\frac{-23}{20}\)

= \(\frac{494}{25}\)

\(a,20,7+1,47:7-0,23\cdot5\)

\(=20,7+0,21-1,15=19,76\)

\(b,1\frac{4}{5}\cdot\frac{7}{8}+\frac{4}{29}:\frac{5}{58}-1\frac{3}{10}\)

\(=\frac{9}{5}\cdot\frac{7}{8}+\frac{4}{29}:\frac{5}{58}-\frac{13}{10}\)

\(=\frac{9\cdot7}{5\cdot8}+\frac{4}{29}\cdot\frac{58}{5}-\frac{13}{10}\)

\(=\frac{63}{40}+\frac{4}{1}\cdot\frac{2}{5}-\frac{13}{10}\)

\(=\frac{63}{40}+\frac{8}{5}-\frac{13}{10}=\frac{63}{40}+\frac{64}{40}-\frac{52}{40}=\frac{63+64-52}{40}=\frac{75}{40}=\frac{15}{8}\)

D=1/2-1/2⁴ +1/2⁷-1/2¹⁰ +.......+1/2⁵⁸

2D =1 - 1/2-1/2⁴ +1/2⁷-1/2¹⁰ +.......+1/2⁵⁷

2D -D =(1 - 1/2-1/2⁴ +1/2⁷-1/2¹⁰ +.......+1/2⁵⁷)-(1/2-1/2⁴ +1/2⁷-1/2¹⁰ +.......+1/2⁵⁸)

D=1-1/2⁵⁸

Dễ thế không làm được à

Trả lời 

B=(3 10/99+4 11/99-58/299).(1/2-4/3-1/6)

=(......................................).0

=0.

1) Đặt \(D=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)

\(\Rightarrow3D=1+\frac{1}{3}+...+\frac{1}{3^{99}}\)

\(\Rightarrow3D-D=\left(1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)

\(\Leftrightarrow2D=1-\frac{1}{3^{100}}\)

\(\Leftrightarrow D=\frac{3^{100}-1}{2\cdot3^{100}}\)

Vậy \(D=\frac{3^{100}-1}{2\cdot3^{100}}\)

2) Ta có: \(\frac{49}{58}\cdot\frac{2^5}{4^2}-\frac{7^2}{-58}\cdot3\)

\(=\frac{49}{58}\cdot2-\frac{49}{58}\cdot3\)

\(=-1\cdot\frac{49}{58}\)

\(=-\frac{49}{58}\)