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\(A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{17\cdot19}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}\)
\(A=\frac{1}{3}-\frac{1}{19}\)
\(A=\frac{16}{57}\)
Dấu "." là dấu nhân nhá ^^
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{17\times19}\right)\times57-2\times\left(x-1\right)=10\)
\(\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}\right)\times57-2\times\left(x-1\right)=10\)
\(\left(\frac{1}{3}-\frac{1}{19}\right)\times57-2\times\left(x-1\right)=10\)
\(\frac{16}{57}\times57-2\times\left(x-1\right)=10\)
\(16-2\times\left(x-1\right)=10\)
\(2\times\left(x-1\right)=16-10\)
\(2\times\left(x-1\right)=6\)
\(x-1=6:2\)
\(x-1=3\)
\(x=3+1\)
\(x=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{15}+2\left(1-\frac{1}{10}\right)\)
\(=\frac{4}{15}+\frac{9}{5}\)
\(=\frac{31}{15}\)
Bài làm :
Ta có :
\(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{13\times15}+\frac{2}{1\times2}+\frac{2}{2\times3}+...+\frac{2}{9\times10}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{15}+2\left(1-\frac{1}{10}\right)\)
\(=\frac{31}{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{2007\times2009}\)
\(C=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2007}-\frac{1}{2009}\)
\(C=\frac{1}{3}-\frac{1}{2009}=\frac{2006}{6027}\)
2/3x5=1/3-1/5
2/5x7=1/5-1/7.........
2/2005x2007=1/2005-1/2007
2/2007x2009=1/2007-1/2009
CỘNG THEO TÙNG VẾ TA ĐƯỢC
C=1/3-1/2009=2006/3X2009
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\right).y=\frac{2}{3}\)
\(\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)y=\frac{2}{3}\)
\(\left(1-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\frac{10}{11}.y=\frac{2}{3}\)
\(y=\frac{2}{3}.\frac{11}{10}\)
\(y=\frac{22}{30}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{x\left(x+2\right)}=\frac{8}{17}\)
\(\Leftrightarrow2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{x\left(x+2\right)}\right)=2.\frac{8}{17}\)
\(\Leftrightarrow\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{x\left(x+2\right)}=\frac{16}{17}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+2}=\frac{16}{17}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{16}{17}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{16}{17}=\frac{1}{17}\)
\(\Rightarrow x+2=17\Rightarrow x=15\)
\(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{17\times19}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}\)
=\(\frac{1}{3}-\frac{1}{19}=\frac{19}{57}-\frac{3}{57}=\frac{19-3}{57}=\frac{16}{57}\)