\(\frac{1}{2^3}\)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

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

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

= \(\frac{2091}{100}+\frac{-23}{20}\)

= \(\frac{494}{25}\)

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

Câu a :

A = 2\(\frac12\) : (- \(\frac12\))^2 - \(\frac{1}{-3}\).(\(\frac{1}{-2}\) - \(\) \(\frac43\) : \(-\frac89\))

A = \(\frac52\) : \(\frac14\) + \(\frac13\).(-\(\frac12\) + \(\frac43\times\frac98\))

A = \(\frac52\times4\) + \(\frac13\).(- \(\frac12+\frac32\))

A = \(10\) + 1/3

A = 30/3 + 1/3

A = 31/3

Câu b:

B = (3\(\frac{10}{99}\) + 4\(\frac{11}{99}\) - \(\frac{58}{229}\)).(\(\frac12\) - \(\frac43\) - \(\frac16\))

B = (7\(\frac{21}{99}\) - \(\frac{58}{229}\))(3/6 - 8/6 - 1/6)

B = (7\(\frac{7}{33}\) - \(\frac{58}{229}\)).(-1)

B = - (\(\frac{238}{33}-\frac{58}{229}\))

B = - (\(54502\)/7557 - 1914/7557)

B = - 52588/7557

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

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

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

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