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\(2014:\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1\frac{2}{5}-\frac{7}{9}+\frac{7}{11}}\cdot\frac{1\frac{1}{6}+0,875-0,7}{\frac{1}{3}+0,25-\frac{1}{5}}\right)\)
\(=2014:\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}\cdot\frac{\frac{7}{6}+\frac{7}{8}-\frac{7}{10}}{\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}\right)\)
\(=2014:\left(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}\cdot\frac{\frac{7}{6}+\frac{7}{8}-\frac{7}{10}}{\frac{2}{6}+\frac{2}{8}-\frac{2}{10}}\right)\)
\(=2014:\left(\frac{2}{7}\cdot\frac{7\left(\frac{1}{6}+\frac{1}{8}-\frac{1}{10}\right)}{2\left(\frac{1}{6}+\frac{1}{8}-\frac{1}{10}\right)}\right)\)
\(=2014:\left(\frac{2}{7}\cdot\frac{7}{2}\right)=2014\)
Ta có : \(1+2=\frac{2.3}{2}\) , \(1+2+3=\frac{3.4}{2}\) ,
\(1+2+3+4=\frac{4.5}{2}\) , ......... , \(1+2+3+4+....+2014=\frac{2014.2015}{2}\)
Suy ra : \(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2014.2015}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2014.2015}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(2\left(\frac{1}{2}-\frac{1}{2015}\right)\)
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2014}\)
\(A=\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+2014\right).2014:2}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2014.2015}\)
\(A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2014.2015}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{2015}\right)\)
\(A=2.\frac{1}{2}-2.\frac{1}{2015}\)
\(A=1-\frac{2}{2015}\)
\(A=\frac{2013}{2015}\)
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2014}\)
\(A=\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+2014\right).2014:2}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2014.2015}\)
\(A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{2015}\right)\)
\(A=2.\frac{1}{2}-2.\frac{1}{2015}\)
\(A=1-\frac{2}{2015}=\frac{2013}{2015}\)
A=2/6+2/12+....+2/4054182
A=2/2.3+2/3.4+...+2/2013.2014
A= (1-2/2014) : 2=503/1007
Bạn xem lại đề câu a) cho rõ lại
Câu b) Tại x=2013 thì B=x2013-(x+1)x2012+(x+1)x2011-(x+1)x2010+...-(x+1)x2+(x+1)x-1
= x2013-x2013-x2012+x2012+x2011-x2011-x2010+..-x3 - x2+x2+x-1
= x-1 = 2012
a)
\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\\A = \left( {\frac{{30}}{{15}} + \frac{5}{{15}} - \frac{6}{{15}}} \right) - \left( {\frac{{105}}{{15}} - \frac{9}{{15}} - \frac{{20}}{{15}}} \right) - \left( {\frac{3}{{15}} + \frac{{25}}{{15}} - \frac{{60}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} - \left( {\frac{{ - 32}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} + \frac{{32}}{{15}}\\A = \frac{{ - 15}}{{15}}\\A = - 1\end{array}\)
b)
\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right)\\A = 2 + \frac{1}{3} - \frac{2}{5} - 7 + \frac{3}{5} + \frac{4}{3} - \frac{1}{5} - \frac{5}{3} + 4\\A = \left( {2 - 7 + 4} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right) + \left( { - \frac{2}{5} + \frac{3}{5} - \frac{1}{5}} \right)\\A = - 1 + 0 + 0 = - 1\end{array}\)
c) \(\left[3\frac{1}{6}-\left(0,06\cdot7\frac{1}{2}+6\frac{1}{4}\cdot0,24\right)\right]:\left(1\frac{2}{3}+2\frac{2}{3}\cdot1\frac{3}{4}\right)\)
\(=\left[\frac{19}{6}-\left(\frac{3}{50}\cdot\frac{15}{2}+\frac{25}{4}\cdot\frac{6}{25}\right)\right]:\left(\frac{5}{3}+\frac{8}{3}\cdot\frac{7}{4}\right)\)
\(=\left[\frac{19}{6}-\left(\frac{9}{20}+\frac{3}{2}\right)\right]:\left(\frac{5}{3}+\frac{14}{3}\right)\)
\(=\left(\frac{19}{6}-\frac{39}{20}\right):\frac{19}{3}=\frac{73}{60}:\frac{19}{3}=\frac{73}{60}\cdot\frac{3}{19}=\frac{73}{380}\)
Bài giải
\(c,\text{ }\left[3\frac{1}{6}-\left(0,06\cdot7\frac{1}{2}+6\frac{1}{4}\cdot0,24\right)\right]\text{ : }\left(1\frac{2}{3}+2\frac{2}{3}\cdot1\frac{3}{4}\right)\)
\(=\left[\frac{19}{6}-\left(\frac{3}{50}\cdot\frac{15}{2}+\frac{25}{4}\cdot\frac{6}{25}\right)\right]\text{ : }\left(\frac{5}{3}+\frac{8}{3}\cdot\frac{7}{4}\right)\)
\(=\left[\frac{19}{6}-\left(\frac{9}{20}+\frac{3}{2}\right)\right]\text{ : }\left(\frac{5}{3}+\frac{56}{12}\right)\)
\(=\left(\frac{19}{6}-\frac{39}{20}\right)\text{ : }\frac{19}{3}\)
\(=\left(\frac{190}{60}-\frac{117}{60}\right)\cdot\frac{3}{19}=\frac{73}{60}\cdot\frac{3}{19}=\frac{73}{380}\)
\(\approx0,4\)