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\(F=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{2006.2009}\)
\(F=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2006}-\frac{1}{2009}\)
\(F=\frac{1}{5}-\frac{1}{2009}\)
\(F=\frac{2004}{10045}\)
\(F=\frac{3}{5.8}+\frac{3}{8.11}+\frac{1}{11.14}+...+\frac{3}{2006.2009}\)
\(F=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2006}-\frac{1}{2009}\)
\(F=\frac{1}{5}-\frac{1}{2009}\)
\(F=0\)
a)1/5.8+1/8.11+1/11.14+...+1/x(x+3)=101/1540
<=>1/3(3/5.8+3/8.11+...+3/x(x+3) =101/1540
<=>1/3(1/5-1/8+1/8-1/11+...+1/x-1/x+3=101/1540
<=>1/5-1/x+3=303/1540<=>1/x+3=1/308
<=>x+3=308<=>x=305
Nguồn CHTT, hihi !
\(E=\frac{15}{11.14}+\frac{15}{14.17}+\frac{15}{17.20}+.....+\frac{15}{74.77}\)
\(=\frac{15}{3}\left(\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+....+\frac{1}{74}-\frac{1}{77}\right)\)
\(=\frac{15}{3}\left(\frac{1}{11}-\frac{1}{77}\right)\)
\(=\frac{30}{77}\)
\(b.\)ghi lại đề nha bn
\(=\frac{2.2306}{1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{230.231}{2}}}\)
\(=\frac{2.2306}{1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{230.231}}\)
\(=\frac{2.2306}{1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{230.231}\right)}\)
\(=\frac{2.2306}{1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{230}-\frac{1}{231}\right)}\)
\(=\frac{2.2306}{1+2.\left(\frac{1}{2}-\frac{1}{231}\right)}\)
\(=\frac{2.2306}{1+1-\frac{2}{231}}\)
\(=\frac{2.2306}{2-\frac{2}{231}}\)
\(=\frac{2.2306}{2\left(1-\frac{1}{231}\right)}\)
\(=\frac{2306}{1-\frac{1}{231}}\)
mình nha bn thanks nhìu <3
a) \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+...+\frac{1}{2016}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\left(\frac{2015}{2}+1\right)+...+\left(\frac{1}{2016}+1\right)+1}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2017}{2}+...+\frac{2017}{2016}+\frac{2017}{2017}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{2017.\left(\frac{1}{2}+...+\frac{1}{2016}+\frac{1}{2017}\right)}\)
\(=\frac{1}{2017}\)
\(\frac{1}{2.15}+\frac{3}{2.11}+\frac{4}{1.11}+\frac{5}{1.2}\)
\(=\frac{1}{30}+\left(\frac{3}{22}+\frac{4}{11}\right)+\frac{5}{2}\)
\(=\frac{1}{30}+\frac{1}{2}+\frac{5}{2}\)
\(=\frac{1}{30}+3\)
\(=\frac{91}{30}\)
a, \(A=\frac{12}{3.7}+\frac{12}{7.11}+...+\frac{12}{195.199}\)
\(=3.\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{195.199}\right)\)
\(=3.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{195}-\frac{1}{199}\right)\)
\(=3.\left(\frac{1}{3}-\frac{1}{199}\right)\)
\(=3.\left(\frac{199}{597}-\frac{3}{597}\right)\)
\(=3.\frac{196}{597}\)
\(=\frac{196}{199}\)
\(\frac{3^2}{2\cdot11}+\frac{3^2}{11\cdot14}+...+\frac{3^2}{197\cdot200}=\frac{3^2}{2\cdot11}+\left(\frac{3^2}{11\cdot14}+...+\frac{3^2}{197\cdot200}\right)\)
\(=\frac{9}{22}+3\left(\frac{3}{11\cdot14}+...+\frac{3}{197\cdot200}\right)=\frac{9}{22}+3\left(\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(=\frac{9}{22}+3\left(\frac{1}{11}-\frac{1}{200}\right)=\frac{9}{22}+3\left(\frac{200}{2200}-\frac{11}{2200}\right)=\frac{9}{22}+3\cdot\frac{189}{2200}\)
\(=3\cdot\left(\frac{3}{22}+\frac{189}{2200}\right)=3\cdot\left(\frac{300}{2200}+\frac{189}{2200}\right)=3\cdot\frac{489}{2200}=\frac{1467}{2200}\)