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\(\left(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\right).y=\frac{2}{3}\)
\(\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)y=\frac{2}{3}\)
\(\left(1-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\frac{10}{11}.y=\frac{2}{3}\)
\(y=\frac{2}{3}.\frac{11}{10}\)
\(y=\frac{22}{30}\)
a ) A = 20,15 x 25,75 + 74,25 x 20,15
A = 20,15 x ( 25,75 + 74,25 )
A = 20,15 x 100
A = 2015
Tính bằng cách thuận tiện nhất
a) A = 20,15 x 25,75 + 74,25 x 20,15
= 20,15 x (25,75 + 74,25)
= 20,15 x 100
= 2015
\(2A=\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{999x1001}\)
\(2A=\frac{3-1}{1x3}+\frac{5-3}{3x5}+\frac{7-5}{5x7}+...+\frac{1001-999}{999x1001}\)
\(2A=\frac{3}{1x3}-\frac{1}{1x3}+\frac{5}{3x5}-\frac{3}{3x5}+\frac{7}{5x7}-\frac{5}{5x7}+...+\frac{1001}{999x1001}-\frac{999}{999x1001}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{999}-\frac{1}{1001}\)
\(2A=1-\frac{1}{1001}=\frac{1000}{1001}\)=> A = 500/1001
\(S.2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(S.2=\frac{1}{1}-\frac{1}{11}\)
\(S.2=\frac{10}{11}\)
\(S=\frac{10}{11}:2\)
\(S=\frac{5}{11}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{15}+2\left(1-\frac{1}{10}\right)\)
\(=\frac{4}{15}+\frac{9}{5}\)
\(=\frac{31}{15}\)
Bài làm :
Ta có :
\(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{13\times15}+\frac{2}{1\times2}+\frac{2}{2\times3}+...+\frac{2}{9\times10}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{15}+2\left(1-\frac{1}{10}\right)\)
\(=\frac{31}{15}\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{9.11}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{11}\right)\)
\(=\frac{1}{2}.\frac{10}{11}\)
\(=\frac{5}{11}\)
\(=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+....+\frac{2}{9\times11}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{11}\right)\)
\(=\frac{1}{2}\times\frac{10}{11}\)
\(=\frac{5}{11}\)
\(A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{17\cdot19}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}\)
\(A=\frac{1}{3}-\frac{1}{19}\)
\(A=\frac{16}{57}\)
Dấu "." là dấu nhân nhá ^^
\(\frac{2}{1x3}+\)\(\frac{2}{3x5}+\)\(\frac{2}{5x7}+\)\(\frac{2}{7x9}+\frac{2}{9x11}+\frac{2}{11x13}\)
= \(\frac{3-1}{1x3}+\frac{5-3}{3x5}+\frac{7-5}{5x7}+\frac{9-7}{7x9}+\frac{11-9}{9x11}\)\(+\frac{13-11}{11x13}\)
= \(\frac{3}{1x3}-\frac{1}{1x3}+\frac{5}{3x5}-\frac{3}{3x5}+\frac{7}{5x7}-\frac{5}{5x7}+\frac{9}{7x9}-\frac{7}{7x9}+\frac{11}{9x11}\)\(-\frac{9}{9x11}\)\(+\frac{13}{11x13}-\frac{11}{11x13}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\)\(\frac{1}{13}\)
= \(1-\frac{1}{13}=\frac{12}{13}\)
Theo cách mk học sẽ suy ra lun
=1/1-1/3+1/3-1/5+1/5-1/7+...+1/2001-1/2003+1/2003-1/2005
=1-1/2005
=2004/2005