\(\frac{1}{3}.\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right]\)

\(\frac{1}{3}\left[\frac{1}{2}-\frac{1}{20}\right]=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}\)

mk đầu tiên đó

=\(\frac{3}{20}=0,15\)

\(\frac{\frac{4}{3}+\frac{4}{7}-\frac{2}{14}}{-1-\frac{3}{7}+\frac{3}{28}}=\frac{4.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{28}\right)}{-3.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{28}\right)}=\frac{-4}{3}\)

\(A=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+.........+\frac{3}{98.101}\right)\)

\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+............+\frac{1}{98}-\frac{1}{101}\right)\)

\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)\)

\(=\frac{4}{3}.\frac{99}{202}\)

\(=\frac{66}{101}\)

\(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{98.101}\) 

\(\frac{4}{3}A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{98.101}\)

\(\frac{4}{3}A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{101}\) 

\(A=\left(\frac{1}{2}-\frac{1}{101}\right).\frac{3}{4}\) 

\(A=\frac{99}{202}.\frac{3}{4}=\frac{297}{808}\)

\(a)\) \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)

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

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)

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

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

Vậy \(A=\frac{2^9-1}{2^9}\)

Chúc bạn học tốt ~ 

A = 1/3^2 + 1/4^2 + 1/5^2 + ... + 1/50^2

1/3^2 = 1/9

1/4^2 < 1/3.4 = 1/3 - 1/4

1/5^2 < 1/4.5 = 1/4 - 1/5

.............................................

1/50^2 < 1/49.50 = 1/49 - 1/50

Cộng vế với vế ta có:

A = 1/3^2+1/4^2+..+1/50^2 = 1/9 + 1/3 - 1/50

A = 4/9 - 1/50 < 4/9

1/3^2 = 1/9

1/4^2 > 1/4.5 = 1/4 - 1/5

1/5^2 > 1/5.6 = 1/5 - 1/6

............................................

1/50^2 > 1/49.50 = 1/49 - 1/50

Cộng vế với vế ta có:

A = 1/3^2+1/4^2+ ...+ 1/50^2 > 1/9+1/4-1/50

A > 1/4 + (1/9 - 1/50)

1/9 > 1/50

1/9 - 1/50 > 0

A > 1/4 + 1/9 - 1/50 > 1/4

Vậy 1/4 < A < 4/9 (đpcm)

\(\frac{4}{2\cdot5}\)+\(\frac{4}{5\cdot8}\)+\(\frac{4}{8\cdot11}\)+.......+\(\frac{4}{8\cdot83}\)=\(\frac{4}{3}\) (\(\frac{3}{2\cdot5}\)+\(\frac{3}{5\cdot8}\) +......+\(\frac{3}{80\cdot83}\) )

=\(\frac{4}{3}\) (\(\frac{1}{2}\) -\(\frac{1}{5}\) +\(\frac{1}{5}\) -\(\frac{1}{8}\) +..........+\(\frac{1}{80}\) -\(\frac{1}{83}\) )

=\(\frac{4}{3}\) (\(\frac{1}{2}\) -\(\frac{1}{83}\) )

=\(\frac{4}{3}\)*\(\frac{81}{166}\) 

=\(\frac{54}{83}\)

A=...

<=>\(A=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{1}{17.20}\right)\)

<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)

<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)

<=>\(A=\frac{1}{6}-\frac{1}{60}< \frac{1}{6}< 1\)

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