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\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{15}+2\left(1-\frac{1}{10}\right)\)
\(=\frac{4}{15}+\frac{9}{5}\)
\(=\frac{31}{15}\)
Bài làm :
Ta có :
\(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{13\times15}+\frac{2}{1\times2}+\frac{2}{2\times3}+...+\frac{2}{9\times10}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{15}+2\left(1-\frac{1}{10}\right)\)
\(=\frac{31}{15}\)
\(A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{17\cdot19}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}\)
\(A=\frac{1}{3}-\frac{1}{19}\)
\(A=\frac{16}{57}\)
Dấu "." là dấu nhân nhá ^^
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
\(\frac{2}{1x3}+\)\(\frac{2}{3x5}+\)\(\frac{2}{5x7}+\)\(\frac{2}{7x9}+\frac{2}{9x11}+\frac{2}{11x13}\)
= \(\frac{3-1}{1x3}+\frac{5-3}{3x5}+\frac{7-5}{5x7}+\frac{9-7}{7x9}+\frac{11-9}{9x11}\)\(+\frac{13-11}{11x13}\)
= \(\frac{3}{1x3}-\frac{1}{1x3}+\frac{5}{3x5}-\frac{3}{3x5}+\frac{7}{5x7}-\frac{5}{5x7}+\frac{9}{7x9}-\frac{7}{7x9}+\frac{11}{9x11}\)\(-\frac{9}{9x11}\)\(+\frac{13}{11x13}-\frac{11}{11x13}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\)\(\frac{1}{13}\)
= \(1-\frac{1}{13}=\frac{12}{13}\)
\(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{17\times19}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}\)
=\(\frac{1}{3}-\frac{1}{19}=\frac{19}{57}-\frac{3}{57}=\frac{19-3}{57}=\frac{16}{57}\)
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
\(\left(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\right).y=\frac{2}{3}\)
\(\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)y=\frac{2}{3}\)
\(\left(1-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\frac{10}{11}.y=\frac{2}{3}\)
\(y=\frac{2}{3}.\frac{11}{10}\)
\(y=\frac{22}{30}\)
\(.A=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{2013}\right)\)
\(A=\frac{1004}{10065}\)
<a class="ptip tipped" data-name="Nguyễn Ngọc Sáng" data-image="http://olm.vn/images/avt/avt424601_60by60.jpg" href="/thanhvien/nguyenngocsang6a" data-uid="125744" data-hasqtip="true" aria-describedby="qtip-2"> Sáng Nguyễn </a>
A=1/5x7+11/7x9+1/9x11+....+1/2011x2013
2xA=2x(1/5x7+1/7x9+1/9x11+...+1/2011x2013
2xA=2/5x7+2/7x9+2/9x11+...+2/2011x2013
2xA=1/5-1/7+1/7-1/9+1/9-1/11+...+1/2011-1/2013
2xA=1/5-1/2013
2xA=2013/10045-5/10045
2xA=2008/10045
A=2008/10045:2
A=2008/10045x1/2
A=1004/10045
Ta có: \(\frac{2}{3\times5}+\frac{2}{5\times7}+\cdots+\frac{2}{2023\times2025}\)
\(=\frac13-\frac15+\frac15-\frac17+\cdots+\frac{1}{2023}-\frac{1}{2025}\)
\(=\frac13-\frac{1}{2025}=\frac{674}{2025}\)
Đặt \(A=\frac{2}{3\times5}+\frac{2}{5\times7}+\cdots+\frac{2}{203\times2025}\)
\(A=\frac13-\frac15+\frac15-\frac17+\cdots+\frac{1}{2023}-\frac{1}{2025}\)
\(A=\frac13-\frac{1}{2025}\)
\(A=\frac{2025}{6075}-\frac{3}{6075}=\frac{2022}{6075}\)
Vậy \(A=\frac{2022}{6075}\)