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ta có: L = \(\frac{7}{3}+\frac{11}{3^2}+\frac{15}{3^3}+...+\frac{403}{3^{100}}\)
<=> \(3L=7+\frac{11}{3}+\frac{15}{3^2} +..+\frac{403}{3^{99}}\)
=> \(3L-L=\left(7+\frac{11}{3}+\frac{15}{3^2}+...+\frac{403}{3^{99}}\right)-\left(\frac{7}{3}+\frac{11}{3^2}+...+\frac{403}{3^{100}}\right)\)
<=> \(2L=7+\left(\frac{11}{3}-\frac{7}{3}\right)+\left(\frac{15}{3^2}-\frac{11}{3^2}\right)+...+\left(\frac{403}{3 ^{99}}-\frac{399}{3^{99}}\right)-\frac{403}{3^{100}}\)
<=> \(2L=7+4\cdot\frac{1}{3}+4\cdot\frac{1}{3^2}+..+4\cdot\frac{1}{3^{99}}-\frac{403}{3^{100}}\)
<=> \(2L=7+4\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\frac{403}{3^{100}}\)
<=>\(2L=7+4\left[\frac{1}{2}\cdot\left(1-\frac{1}{3^{99}}\right)\right]-\frac{403}{3^{100}}\)
<=> \(2L=7+2-\frac{2}{3^{99}}-\frac{403}{3^{100}}\)
<=> \(L=3,5+1-\frac{1}{3^{99}}-\frac{403}{2\cdot3^{100}}\)
<=> \(L=4,5-\frac{1}{3^{99}}-\frac{403}{2\cdot3^{100}}
\(A=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}\)
\(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}\)
\(=\frac{2}{3}\)
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 \(F=1+\frac{1}{3}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3F=3+1+...+\frac{1}{3^{97}}\)
\(\Rightarrow3F-F=\left(3+...+\frac{1}{3^{97}}\right)-\left(1+...+\frac{1}{3^{98}}\right)\)
\(\Rightarrow2F=3-\frac{1}{3^{98}}< 3\)
\(\Rightarrow F< \frac{3}{2}\)
\(\Rightarrow2A< 4+\frac{3}{2}\)
\(\Rightarrow2A< \frac{11}{2}\)
\(\Rightarrow A< \frac{11}{4}\left(đpcm\right)\)
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 )
\(\Rightarrow2D< 6\)
\(\Rightarrow D< 3\)
\(\Rightarrow2B< 11+3\)
\(\Rightarrow2B< 14\)
\(\Rightarrow B< 7\left(đpcm\right)\)
2: \(=\dfrac{0.8}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\dfrac{71}{75}\cdot\dfrac{7}{4}}{\dfrac{119}{36}\cdot\dfrac{36}{17}}\)
\(=\dfrac{4}{5}\cdot\dfrac{5}{3}+\dfrac{71}{300}=\dfrac{471}{300}=\dfrac{157}{100}\)
3: \(=\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\)
=2/7-2/7=0
\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)
\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)
\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)
\(\Rightarrow-1\le x< 6\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)
Bài b tương tự
\(A=\left(\frac{5}{17}+\frac{12}{17}\right)+\left(-\frac{20}{31}-\frac{11}{31}\right)+-\frac{4}{9}\)
\(A=1+\left(-1\right)+\frac{-4}{9}\)
\(A=0+\frac{-4}{9}=\frac{-4}{9}\)
\(B=\left(\frac{-3}{7}+\frac{-4}{7}\right)+\left(\frac{7}{15}+\frac{8}{15}\right)+\frac{-2}{3}\)
\(B=-1+1+\frac{-2}{3}\)
\(B=\frac{-2}{3}\)