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a) 4/ 3x7 + 4/7x11+ 4/11x15+...+ 4/107x111
=1/3-1/7+ 1/7-1/11+ 1/11- 1/15+...+1/107 - 1/111
= 1/3-1/111
=12/37
\(b,\frac{3^2}{8\cdot11}+\frac{3^2}{11\cdot14}+\frac{3^2}{14\cdot17}+...+\frac{3^2}{197\cdot200}\)
\(=3\left(\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+\frac{3}{14\cdot17}+...+\frac{3}{197\cdot200}\right)\)
\(=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(=3\left(\frac{1}{8}-\frac{1}{200}\right)\)
\(=3\cdot\frac{3}{25}=\frac{9}{25}\)
\(B=5\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{51}\right)=\)\(\frac{250}{51}\)
\(B=5\left(\frac{1}{1}-\frac{1}{51}\right)=\frac{250}{51}\)
\(B=5\left(\frac{51}{51}-\frac{1}{51}\right)=\frac{250}{51}\)
\(B=5.\frac{50}{51}=\frac{250}{51}\)
\(B=\frac{250}{51}=\frac{250}{51}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{2}-\frac{1}{50}\)
\(=\frac{12}{25}\)
\(B=\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{23.27}\)
\(=\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{23.27}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{27}\right)\)
\(=\frac{1}{4}.\frac{8}{27}=\frac{2}{27}\)
Sửa đề \(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+....+\frac{4}{107\cdot111}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{107}-\frac{1}{111}\)
\(=\frac{1}{3}-\frac{1}{111}=\frac{12}{37}\)
Ta thấy \(\frac{1}{3}-\frac{1}{7}=\frac{7-3}{3.7}=\frac{4}{3.7}\)
\(\frac{1}{7}-\frac{1}{11}=\frac{11-7}{7.11}=\frac{4}{7.11}\)
..........................
\(\frac{1}{1023}-\frac{1}{1027}=\frac{1027-1023}{1023.1027}=\frac{4}{1023.1027}\)
=> \(\frac{4}{3.7}+\frac{4}{7.11}+....+\frac{4}{1023.1027}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{1023}-\frac{1}{1027}\)
=> =\(\frac{1}{3}-\frac{1}{1027}=\frac{1024}{3.1027}\)
Ta có: \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{1023.1027}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{1023}-\frac{1}{1027}\)
\(=\frac{1}{3}-\frac{1}{1027}=\frac{1024}{3081}\)
\(\frac{1}{7}\)-\(\frac{3}{4}\): \(x\)=\(\frac{3}{5}\)
\(\frac{3}{4}\):\(x\)=\(\frac{1}{7}\)-\(\frac{3}{5}\)
\(\frac{3}{4}\):\(x\)=\(\frac{-16}{35}\)
\(x\)= \(\frac{3}{4}\):\(\frac{-16}{35}\)
\(x\)= \(\frac{3}{4}\). \(\frac{-35}{16}\)
\(x\)=\(\frac{-105}{64}\)
Vậy \(x\)=\(\frac{-105}{64}\)
\(\frac{\frac{2}{3}+\frac{2}{5}-\frac{2}{9}}{\frac{4}{3}+\frac{4}{5}-\frac{4}{9}}=\frac{1}{2}\)
\(4\frac{2}{5}+5\frac{6}{9}+2\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=4+\frac{2}{5}+5+\frac{6}{9}+2+\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\left(4+5+2\right)+\left(\frac{2}{5}+\frac{3}{5}\right)+\left(\frac{6}{9}+\frac{1}{3}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)\)
\(=11+1+1+1\)
\(=14\)
\(=\frac{22}{5}+\frac{17}{3}+\frac{11}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}\)
\(=\left(\frac{22}{5}+\frac{3}{5}\right)+\left(\frac{11}{4}+\frac{1}{4}\right)+\left(\frac{17}{3}+\frac{1}{3}\right)\)
\(=5+3+6=14\)
=\(4\left(\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{1023.1027}\right)\)
=\(4\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{1023}-\frac{1}{1027}\right)\)
=\(4\left(\frac{1}{3}-\frac{1}{1027}\right)\)
=\(\frac{1352}{1017}\)
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