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Giải thích thêm: ta thấy \(\frac{1}{2^2}>\frac{1}{100}\),...,\(\frac{1}{10^2}=\frac{1}{100}\)=> từ \(\frac{1}{2^2}\)đến \(\frac{1}{10^2}\)có 5 cặp
\(\frac{1}{12^2}< \frac{1}{100}\),...,\(\frac{1}{100^2}< \frac{1}{100}\)=> từ \(\frac{1}{12^2}\)đến \(\frac{1}{100^2}\)có 45 cặp
=> 45>5 => tổng < 1/2 (kết hợp với cái kia nx thì bn mới hiểu)
\(\frac{2}{2.3}\) + \(\frac{2}{3.4}\) + \(\frac{2}{4.5}\) + .......+ \(\frac{2}{x.\left(x+1\right)}\) = \(\frac{2017}{2019}\)
2 . ( \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\) + .......+ \(\frac{1}{x+1}\) ) = \(\frac{2017}{2019}\)
2 . ( \(\frac{1}{2}\) - \(\frac{1}{x+1}\) ) = \(\frac{2017}{2019}\)
\(\frac{1}{2}\) - \(\frac{1}{x+1}\) = \(\frac{2017}{2019}\) : 2
\(\frac{1}{2}\) - \(\frac{1}{x+1}\) = \(\frac{2017}{4038}\)
\(\frac{1}{x+1}\) = \(\frac{1}{2}\) - \(\frac{2017}{4038}\)
\(\frac{1}{x+1}\) = \(\frac{1}{2019}\)
<=> x + 1 = 2019 => x = 2018
vậy x = 2018
\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2017}{4038}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2019}\)
\(\Rightarrow x+1=2019\)
\(\Leftrightarrow x=2018\)
Vậy \(x=2018\)
1. a) \(\frac{-2}{7}+\frac{15}{23}+\frac{\left(-15\right)}{17}+\frac{4}{19}+\frac{8}{23}\)
\(=\left(\frac{-2}{7}+\frac{-5}{7}\right)+\left(\frac{15}{23}+\frac{8}{23}\right)+\frac{4}{19}\)
\(=\left(-1\right)+1+\frac{4}{19}\)
\(=0+\frac{4}{19}=\frac{4}{19}\)
b) \(\frac{7}{19}\cdot\frac{8}{11}+\frac{7}{19}\cdot\frac{3}{11}+\frac{12}{19}\)
\(=\frac{7}{19}\cdot\left(\frac{8}{11}+\frac{3}{11}\right)+\frac{12}{19}\)
\(=\frac{7}{19}\cdot1+\frac{12}{19}\)
\(=\frac{7}{19}+\frac{12}{19}=\frac{19}{19}=1\)
2. a) \(\frac{1}{3}+\frac{\left(-2\right)}{16}-\frac{7}{14}\)
\(=\frac{5}{24}-\frac{1}{2}\)
\(=-\frac{7}{24}\)
b) \(11\frac{3}{13}-2\frac{4}{7}+5\frac{3}{13}\)
\(=\left(11-2+5\right)+\frac{3}{13}-\frac{4}{7}+\frac{3}{13}\)
\(=14+\left(-\frac{10}{91}\right)\)
\(=-14\frac{10}{91}\)
c) \(0,7\cdot2\frac{2}{3}\cdot20\cdot0,375\cdot\frac{5}{28}\)
\(=\frac{7}{10}\cdot\frac{8}{3}\cdot20\cdot\frac{3}{8}\cdot\frac{5}{28}\)
\(=\left(\frac{7}{10}\cdot\frac{5}{28}\right)\cdot\left(\frac{8}{3}\cdot\frac{3}{8}\right)\cdot20\)
\(=\frac{1}{8}\cdot1\cdot20\)
\(=\frac{20}{8}=\frac{5}{2}\)
d) \(\frac{6}{7}+\frac{5}{7}:5-\frac{8}{9}\)
\(=\frac{6}{7}+\frac{1}{7}-\frac{8}{9}\)
\(=1-\frac{8}{9}\)
\(=\frac{1}{9}\)
~Học tốt~
hỏi chị google ấy
A= \(\frac{1}{31}.\left[\frac{5}{31}\left(9-\frac{1}{2}\right)-\frac{17}{2}\left(4+\frac{1}{5}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
= \(\frac{1}{31}.\left(\frac{5}{31}.\frac{17}{2}-\frac{17}{2}.\frac{21}{5}\right)+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{1}{31}.\left[\frac{17}{2}.\left(\frac{5}{31}-\frac{21}{5}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{1}{31}.\left[\frac{17}{2}.\left(\frac{-626}{155}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{1}{31}.\left(\frac{-5321}{155}\right)+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{-5321}{4805}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{-5321}{4805}+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{30.31}\)
=\(\frac{-5321}{4805}+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{30}-\frac{1}{31}\)
=\(\frac{-5321}{4805}+\frac{1}{1}-\frac{1}{31}\)
=\(\frac{-5321}{4805}+\frac{30}{31}\)
=\(\frac{-671}{4805}\)