\(\frac{3}{5x7}\)+ \(\frac{3}{7x9}\) + \(\frac{3}{9x11}\) + ............................+ \(\frac{3}{63x65}\)
Tìm y biết:
\(\left(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\right)\cdot y=\frac{2}{3}\)
\(\left(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\right).y=\frac{2}{3}\)
\(\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)y=\frac{2}{3}\)
\(\left(1-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\frac{10}{11}.y=\frac{2}{3}\)
\(y=\frac{2}{3}.\frac{11}{10}\)
\(y=\frac{22}{30}\)
\(\left(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\right).y=\frac{2}{3}\)
\(\frac{10}{11}.y=\frac{2}{3}\)
\(y=\frac{11}{15}\)
\(\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).y=\frac{2}{3}\)
\(\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{11-9}{9.11}\right).y=\frac{2}{3}\)
\(\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\left(\frac{1}{1}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\frac{10}{11}.y=\frac{2}{3}\)
\(y=\frac{2}{3}:\frac{10}{11}\)
\(y=\frac{11}{15}\)
\(\frac{3}{5x7}\)+ \(\frac{3}{7x9}\)+ \(\frac{3}{9x11}\)+ ... + \(\frac{3}{2013x2015}\)
Đây là bài tính nhanh, giúp mk nhé
\(\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+\frac{3}{9\cdot11}+...+\frac{3}{2013\cdot2015}\)
\(=\frac{3}{2}\cdot\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}\cdot\left(\frac{1}{5}-\frac{1}{2015}\right)=\frac{3}{2}\cdot\frac{402}{2015}-\frac{603}{2015}\)
Vậy \(\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+\frac{3}{9\cdot11}+...+\frac{3}{2013\cdot2015}=\frac{603}{2015}\)
3/5.7 + 3/7.9 + 3/9.11 + ... 3/2013.2015
= 3/2.( 2/5.7 + 2/7.9 + 2/9.11 + ... + 2/2013.2015)
= 3/2. ( 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + ... + 1/2013 - 1/2015)
~~~ SAU ĐÓ BẠN GẠCH ĐI NHỮNG PHÂN SỐ GIỐNG NHAU NHÁ ~~~
= 3/2. ( 1/5 - 1/2015)
= 3/2. 2010/10075
= 603/4030
Mk chắc chắn cách làm đúng đó!!!
\(\frac{3}{5\text{×}7}+\frac{3}{7\text{×}9}+.......+\frac{3}{2013\text{×}2015}\)
\(=\frac{3}{2}\text{×}\left(\frac{2}{5\text{×}7}+\frac{2}{7\text{×}9}+........+\frac{2}{2013\text{×}2015}\right)\)
\(=\frac{3}{2}\text{×}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+......+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}\text{×}\left(\frac{1}{5}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}\text{×}\frac{402}{2015}\)
\(=\frac{603}{2015}\)
\(\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}+...+\frac{2}{13x15}\)
\(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)
\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{13}-\frac{1}{15}\)
\(=\frac{1}{5}-\frac{1}{15}=\frac{3}{15}-\frac{1}{15}=\frac{3-1}{15}=\frac{2}{15}\)
P/s : Dấu chấm là nhân nhé!
2/5x7+2/7x9+2/9x11+...+2/13x15
=1/5x7+1/7x9+1/9x11+...+1/13x15
=1/5-1/15
=3/15-1/15
=2/15
Tính bằng cách thuận tiện:
\(\frac{3}{5x3}\)+ \(\frac{3}{5x7}\)+\(\frac{3}{7x9}\)+\(\frac{3}{9x11}\)+\(\frac{3}{11x13}\)
#(các bạn giúp mik nha, mk sẽ tích đúng)
\(\frac{3}{5×3}+\frac{3}{5×7}+\frac{3}{7×9}+\frac{3}{9×11}+\frac{3}{11×13}\)
\(=\frac{3}{2}×\left(\frac{2}{3×5}+\frac{2}{5×7}+\frac{2}{7×9}+\frac{2}{9×11}+\frac{2}{11×13}\right)\)
\(=\frac{3}{2}×\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{3}{2}×\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=\frac{3}{2}×\left(\frac{13}{39}-\frac{3}{39}\right)\)
\(=\frac{3}{2}×\frac{10}{39}\)
\(=\frac{5}{13}\)
\(\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+\frac{3}{11.13}\)
\(=\frac{3}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{ 1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11} +\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=\frac{3}{2}.\frac{10}{39}\)
\(=\frac{15}{39}\)
\(\left(\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+\frac{1}{9x11}\right)\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{9.11}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{11}\right)\)
\(=\frac{1}{2}.\frac{10}{11}\)
\(=\frac{5}{11}\)
\(=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+....+\frac{2}{9\times11}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{11}\right)\)
\(=\frac{1}{2}\times\frac{10}{11}\)
\(=\frac{5}{11}\)
\(B=\frac{4}{3x5}-\frac{6}{5x7}+\frac{8}{7x9}-\frac{10}{9x11}+\frac{12}{11x13}-...+\frac{100}{99x101}\)
Tính
\(\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+\frac{1}{9x11}\)
\(S.2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(S.2=\frac{1}{1}-\frac{1}{11}\)
\(S.2=\frac{10}{11}\)
\(S=\frac{10}{11}:2\)
\(S=\frac{5}{11}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(=\frac{1}{1}-\frac{1}{11}\)
\(=\frac{10}{11}\)
\(\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}+\frac{2}{11x13}+\frac{2}{13x15}\)Tính
\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}+\frac{2}{13\cdot15}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(=\frac{1}{3}-\frac{1}{15}\)
\(=\frac{4}{15}\)
Chúc bn hok giỏi !!!!!!!!! ^_^
1, Tìm x
\(\frac{1}{5x7}+\frac{1}{7x9}+\frac{1}{9x11}+.....+\frac{1}{2009x2011}+\frac{1}{x}=\frac{1}{5}x0,5\)
\(\frac{1}{5.7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+...+\frac{1}{2009\cdot2011}+\frac{1}{x}=\frac{1}{5}\cdot0,5\)
\(=\frac{7-5}{5\cdot7}+\frac{9-7}{7\cdot9}+\frac{11-9}{9\cdot11}+...+\frac{2011-2009}{2009\cdot2011}+\frac{1}{x}=\frac{1}{10}\)
\(=\left[\frac{1}{2}\cdot\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{2009}-\frac{1}{2011}\right)\right]+\frac{1}{x}=\frac{1}{10}\)
\(=\left[\frac{1}{2}\cdot\left(\frac{1}{5}-\frac{1}{2011}\right)\right]+\frac{1}{x}=\frac{1}{10}\)
\(=\left(\frac{1}{2}\cdot\frac{2006}{10055}\right)+\frac{1}{x}=\frac{1}{10}\)
\(=\frac{1003}{10055}+\frac{1}{x}=\frac{1}{10}\)
\(\Rightarrow\frac{1}{x}=\frac{1}{10}-\frac{1003}{10055}\)
\(\frac{1}{x}=\frac{1}{4022}\)
\(\Rightarrow x=1\div\frac{1}{4022}=4022\)