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19 tháng 2 2019

Ta có \(I=\frac{11}{3}+\frac{17}{3^2}+...+\frac{605}{3^{100}}\left(1\right)\)

\(\Leftrightarrow3I=11+\frac{17}{3}+\frac{23}{3^2}+...+\frac{605}{3^{99}}\left(2\right)\)

Lấy \(\left(2\right)trừ\left(1\right)\)ta có

\(3I-I=11+\frac{6}{3}+\frac{6}{3^2}+...+\frac{6}{3^{99}}-\frac{605}{3^{100}}\)

\(\Leftrightarrow2I=11+6\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\frac{605}{3^{100}}\)

Xét \(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\left(3\right)\)

\(\Leftrightarrow3A=1+\frac{1}{3}+...+\frac{1}{3^{99}}\left(4\right)\)

Lấy\(\left(4\right)-\left(3\right)\)ta có

\(2A=1-\frac{1}{3^{100}}\)

\(\Leftrightarrow6A=3-\frac{1}{3^{99}}\)

Khi đó \(2I=11+3-\frac{1}{3^{99}}-\frac{605}{3^{100}}\)

\(\Leftrightarrow2I=14-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)

\(\frac{1}{3^{99}}+\frac{605}{3^{100}}>0\)

\(\Rightarrow2I< 14\)

\(\Leftrightarrow I< 7\left(đpcm\right)\)

27 tháng 7 2019

a) \(A=\frac{4}{3}+\frac{7}{3^2}+\frac{10}{3^3}+...+\frac{301}{3^{100}}\)

\(\Rightarrow3A=4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{301}{3^{100}}\)

\(\Rightarrow3A-A=\left(4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{301}{3^{99}}\right)-\left(\frac{4}{3}+\frac{7}{3^2}+...+\frac{301}{3^{100}}\right)\)

\(\Rightarrow2A=4+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{301}{3^{100}}\)

Đặt \(F=1+\frac{1}{3}+...+\frac{1}{3^{98}}\)

\(\Rightarrow3F=3+1+...+\frac{1}{3^{97}}\)

\(\Rightarrow3F-F=\left(3+...+\frac{1}{3^{97}}\right)-\left(1+...+\frac{1}{3^{98}}\right)\)

\(\Rightarrow2F=3-\frac{1}{3^{98}}< 3\)

\(\Rightarrow F< \frac{3}{2}\)

\(\Rightarrow2A< 4+\frac{3}{2}\)

\(\Rightarrow2A< \frac{11}{2}\)

\(\Rightarrow A< \frac{11}{4}\left(đpcm\right)\)

27 tháng 7 2019

2. \(B=\frac{11}{3}+\frac{17}{3^2}+\frac{23}{3^3}+...+\frac{605}{3^{100}}\)

\(\Rightarrow3B=11+\frac{17}{3}+\frac{23}{3^2}+...+\frac{605}{3^{99}}\)

\(\Rightarrow3B-B=\left(11+...+\frac{605}{3^{99}}\right)-\left(\frac{11}{3}+...+\frac{605}{3^{100}}\right)\)

\(\Rightarrow2B=11+2+\frac{2}{3}+...+\frac{2}{3^{98}}-\frac{605}{3^{100}}\)

Đặt \(D=2+\frac{2}{3}+...+\frac{2}{3^{98}}\)

\(\Rightarrow3D=6+2+...+\frac{2}{3^{97}}\)

\(\Rightarrow2D=6-\frac{2}{3^{98}}< 6\)( làm tắt )

\(\Rightarrow2D< 6\)

\(\Rightarrow D< 3\)

\(\Rightarrow2B< 11+3\)

\(\Rightarrow2B< 14\)

\(\Rightarrow B< 7\left(đpcm\right)\)

19 tháng 2 2019

Ta có \(H=\frac{7}{3}+\frac{13}{3^2}+...+\frac{605}{3^{100}}\)

\(\Leftrightarrow3H=7+\frac{13}{3}+...+\frac{605}{3^{99}}\)

\(\Rightarrow2H=7+\frac{6}{3}+\frac{6}{3^2}+...+\frac{6}{3^{99}}-\frac{605}{3^{100}}\)

\(\Leftrightarrow2H=7+6\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\frac{605}{3^{100}}\)

Mà \(6\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)=3-\frac{1}{3^{99}}\)

\(\Rightarrow2H=7+3-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)

\(\Leftrightarrow2H=10-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)

\(\frac{1}{3^{99}}+\frac{605}{3^{100}}>0\)

\(\Rightarrow2H< 10\)

\(\Leftrightarrow H< 5\left(1\right)\)

Ta có \(2H=10-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)

\(\frac{1}{3^{97}}+\frac{605}{3^{98}}< 22\)

hay\(\frac{1}{3^{99}}+\frac{605}{3^{98}}< \frac{22}{9}\)

\(\Rightarrow2H>10-\frac{22}{9}=\frac{68}{9}=2\cdot\left(3+\frac{7}{9}\right)\)

\(\Rightarrow H>3+\frac{7}{9}\left(2\right)\)

Từ \(\left(1\right)\left(2\right)\Rightarrowđpcm\)

 

Sai r

 

1 tháng 8 2018

a, \(\frac{3}{8}+\frac{11}{13}-\frac{9}{13}\)

  =\(\frac{3}{8}+\frac{2}{13}\)

  =\(\frac{55}{104}.\)

b, \(\frac{2}{7}.\left(\frac{5}{9}+\frac{4}{9}\right)+\frac{2}{7}\)

  =\(\frac{2}{7}.\frac{9}{9}+\frac{2}{7}\)

  =\(\frac{2}{7}+\frac{2}{7}\)

  =\(\frac{4}{7}\)

c, \(\frac{3}{11}.\left(\frac{3}{5}-\frac{5}{3}\right)-\frac{3}{10}.\left(\frac{1}{3}-\frac{2}{5}\right)\)

  =\(\frac{3}{11}.-\frac{16}{15}-\frac{3}{10}.-\frac{1}{15}\)

  =\(-\frac{16}{55}--\frac{1}{50}\)

  =\(-\frac{149}{550}.\)

d, \(\frac{-3}{4}.\frac{11}{23}+\frac{3}{23}.\frac{31}{17}-\frac{3}{17}.\frac{19}{23}\)

  =\(-\frac{33}{92}+\frac{93}{391}-\frac{57}{391}\)

  =\(-\frac{417}{1564}\)

e, \(\frac{3}{17}.\frac{11}{23}+\frac{3}{23}.\frac{31}{17}-\frac{3}{17}.\frac{19}{23}\)

  =\(\frac{33}{391}+\frac{93}{391}--\frac{254}{391}\)

  =\(\frac{380}{391}.\)

g, \(\frac{3}{7}.\frac{-5}{12}+\frac{11}{17}:\frac{5}{-12}\)

  =\(-\frac{5}{28}+-\frac{132}{85}\)

  = \(-1.731512605.\)

k cho mình nha làm mỏi tay quá ,.....................kết bạn với mình nha.......................

5 tháng 8 2018

THANK  Ngô Bùi Hoa  làm cho mình bài 2 với 

2 tháng 7 2016

\(M=\frac{2.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{23}\right)}{-5.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{23}\right)}+\frac{\frac{1}{17}-\frac{1}{23}+\frac{1}{31}}{3.\left(\frac{1}{17}-\frac{1}{23}+\frac{1}{31}\right)}=-\frac{2}{5}+\frac{1}{3}=\frac{1}{15}.\)