Cho H = \(\frac{7}{3}+\frac{13}{3^2}+\frac{19}{3^3}+.....+\frac{605}{3^{100}}\)
CMR \(3\frac{7}{9}< H< 5\)
K
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NA
0
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 \(2I=11+3-\frac{1}{3^{99}}-\frac{605}{3^{100}}\)
\(\Leftrightarrow2I=14-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)
Vì\(\frac{1}{3^{99}}+\frac{605}{3^{100}}>0\)
\(\Rightarrow2I< 14\)
\(\Leftrightarrow I< 7\left(đpcm\right)\)
25 tháng 4 2017
\(-\frac{3}{5}.\frac{5}{7}+-\frac{3}{5}.\frac{3}{7}+-\frac{3}{5}.\frac{6}{7}=-\frac{3}{5}\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)=-\frac{3}{5}.2=-\frac{6}{5}\)
\(\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}-\frac{4}{3}=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}\right)-\frac{4}{3}=\frac{1}{3}.2-\frac{4}{3}=\frac{2}{3}-\frac{4}{3}=-\frac{2}{3}\)
\(\frac{4}{19}.\frac{-3}{7}+-\frac{3}{7}.\frac{15}{19}+\frac{5}{7}=-\frac{3}{7}\left(\frac{4}{19}+\frac{15}{19}\right)+\frac{5}{7}=-\frac{3}{7}+\frac{5}{7}=\frac{2}{7}\)
\(\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)=\frac{5}{9}\)
25 tháng 4 2017
\(a,6\frac{4}{5}-\left(1\frac{2}{3}+3\frac{4}{5}\right)=6\frac{4}{5}-1\frac{2}{3}-3\frac{4}{5}=6\frac{4}{5}-3\frac{4}{5}-1\frac{2}{3}=3-1\frac{2}{3}=\frac{4}{3}\)
\(b,6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)=6\frac{5}{7}-2\frac{5}{7}-1\frac{3}{4}=\frac{9}{4}\)
\(c,7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)=7\frac{5}{9}-3\frac{5}{9}-2\frac{3}{4}=4-2\frac{3}{4}=\frac{5}{4}\)
mk nghĩ là phần d như thế này cơ \(7\frac{5}{11}\left(2\frac{3}{7}+3\frac{5}{11}\right)\)
\(7\frac{5}{11}-\left(2\frac{3}{7}+3\frac{5}{11}\right)=7\frac{5}{11}-3\frac{5}{11}-2\frac{3}{7}=4-2\frac{3}{7}=\frac{11}{7}\)
TA
0
25 tháng 9 2018
1.\(\left(-\frac{6}{5}+\frac{6}{16}-\frac{6}{23}\right):\left(\frac{9}{5}-\frac{9}{16}+\frac{9}{23}\right)\)
\(=6\left(-\frac{1}{5}+\frac{1}{16}-\frac{1}{23}\right):\left(-9\right)\left(\frac{-1}{5}+\frac{1}{16}-\frac{1}{23}\right)\)
\(=6:\left(-9\right)=-\frac{2}{3}\)
2. \(\frac{\frac{3}{7}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}+\frac{0.5-\frac{1}{3}+\frac{1}{4}}{-\frac{3}{2}+1-\frac{3}{4}}\)
\(=\frac{3\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{-3\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}-\frac{1}{3}\)
\(=\frac{9}{13}-\frac{5}{15}=\frac{4}{15}\)
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 \(6\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)=3-\frac{1}{3^{99}}\)
\(\Rightarrow2H=7+3-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)
\(\Leftrightarrow2H=10-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)
Vì\(\frac{1}{3^{99}}+\frac{605}{3^{100}}>0\)
\(\Rightarrow2H< 10\)
\(\Leftrightarrow H< 5\left(1\right)\)
Ta có \(2H=10-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)
Mà\(\frac{1}{3^{97}}+\frac{605}{3^{98}}< 22\)
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 \(\left(1\right)\left(2\right)\Rightarrowđpcm\)