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15 tháng 4 2017

\(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)

\(=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)

\(=2\times\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...\dfrac{1}{240}\right)\)

\(=2\times\left(\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+...+\dfrac{1}{15\times16}\right)\)

\(=2\times\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(=2\times\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)

\(=\dfrac{3}{8}\)

15 tháng 4 2017

=2/20+2/30+2/42+.....+2/240

=2/4.5+2/5.6+2/6.7+.....+2/15.16

=1/2[1/4.5+1/5.6+1/6.7+.....+1/15.16]

=1.2[1/4-1/5+1/5-1/6+.....+1/15-1/16]

=1/2[1/4-1/16]

=1/2.3/16

=3/32

3 tháng 4 2018

\(B=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)

\(\Leftrightarrow B=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)

\(\Leftrightarrow B=2\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{240}\right)\)

\(\Leftrightarrow B=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)

\(\Leftrightarrow B=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(\Leftrightarrow B=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{3}{8}\)

\(\dfrac{3}{8}< \dfrac{1}{2}\)

\(\Rightarrow B< \dfrac{1}{2}\left(ĐPCM\right)\)

16 tháng 4 2017

\(\dfrac{x}{2008}-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{21}-...-\dfrac{1}{120}=\dfrac{5}{8}\)

\(\dfrac{x}{2008}-\dfrac{2}{20}-\dfrac{2}{30}-\dfrac{2}{42}-...-\dfrac{2}{240}=\dfrac{5}{8}\)

\(\dfrac{x}{2008}-\left(\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\right)=\dfrac{5}{8}\)

\(\dfrac{x}{2008}-2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)=\dfrac{5}{8}\)

\(\dfrac{x}{2008}-2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)=\dfrac{5}{8}\)

\(\dfrac{x}{2008}-2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{5}{8}\)

\(\dfrac{x}{2008}-2.\dfrac{3}{16}=\dfrac{5}{8}\)

\(\dfrac{x}{2008}-\dfrac{3}{8}=\dfrac{5}{8}\)

\(\dfrac{x}{2008}=\dfrac{5}{8}+\dfrac{3}{8}\)

\(\dfrac{x}{2008}=1=\dfrac{2008}{2008}\)

\(\Rightarrow x=2008\)

17 tháng 3 2017

b,\(\dfrac{1}{3.5}+\dfrac{1}{5.7}\)\(+\dfrac{1}{7.9}+....+\dfrac{1}{\left(2x+1\right).\left(2x+3\right)}=\dfrac{15}{93}\)

\(\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2x+1}-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)

\(\left[\dfrac{1}{3}+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+....+\left(\dfrac{1}{2x+1}-\dfrac{1}{2x+1}\right)-\dfrac{1}{2x+3}\right].\dfrac{1}{2}=\dfrac{15}{93}\)

\(\left(\dfrac{1}{3}+0+0+...+0-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)

\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{15}{93}:\dfrac{1}{2}\)

\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{10}{31}\)

\(\dfrac{1}{2x+3}=\dfrac{1}{3}-\dfrac{10}{31}\)

\(\dfrac{1}{2x+3}=\dfrac{1}{93}\)

\(\Rightarrow2x+3=93\)

\(2x=93-3=90\)

\(\Rightarrow x=90:2=45\)

19 tháng 3 2017

Cảm ơn bạn

11 tháng 3 2017

\(A=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)

\(A=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)

\(A=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{15.16}\)

\(A=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)

\(A=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(A=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)

\(A=2.\dfrac{3}{16}\)

\(A=\dfrac{3}{8}\)

11 tháng 3 2017

\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)

\(B=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)

\(B=\dfrac{1}{3}-\dfrac{1}{111}\)

\(B=\dfrac{12}{37}\)

10 tháng 4 2018

A= \(\dfrac{1}{1.2}\)+ \(\dfrac{1}{2.3}\)+ \(\dfrac{1}{3.4}\)+ \(\dfrac{1}{4.5}\)+ \(\dfrac{1}{5.6}\)

= 1-\(\dfrac{1}{2}\)+ \(\dfrac{1}{2}\)- \(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\)- \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)- \(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)- \(\dfrac{1}{6}\)

= 1 - \(\dfrac{1}{6}\)= \(\dfrac{5}{6}\)

mk chỉ bt làm câu 1 thôi ak

mong bn thông cảmthanghoa

2 tháng 7 2017

\(A=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}=2\times\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{240}\right)\)

\(A=2\times\left(\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+....+\dfrac{1}{15\times16}\right)\)

\(A=2\times\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(A=2\times\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{3}{8}\)

2 tháng 7 2017

b) cậu đi tìm số sốm hạng là : \(\left(2010-1\right):1+1=2010\)

\(\Rightarrow\)số cặp trong phép tính là : \(2010:2=1005\)(cặp)

\(\Rightarrow B=1-2+3-4+...+2009-2010\)(1005 cặp)

\(\Rightarrow\left(1-2\right)+\left(3-4\right)+...+\left(2009-2010\right)\)

\(\Rightarrow B=\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)(1005 số -1)

\(\Rightarrow B=\left(-1\right).1005\)

\(\Rightarrow B=\left(-1005\right)\)

cậu tik cho mik nhé!!!

10 tháng 4 2017

A =\(2.\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+......+\dfrac{1}{156}\right)\)

A =\(2.\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+..........+\dfrac{1}{12.13}\right)\)

A =2.\(\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)

A=\(2.\dfrac{10}{39}=\dfrac{20}{39}\)

10 tháng 4 2017

tớ làm hơi gọn nên có gì kho hiểu thì nói tớ

30 tháng 3 2017

=\(\dfrac{1}{3.2}+\dfrac{1}{2.5}+\dfrac{1}{5.3}+\dfrac{1}{3.7}+\dfrac{1}{7.4}+\dfrac{1}{4.9}+\dfrac{1}{9.5}\)=\(\dfrac{1}{3}+\dfrac{1}{5}\)

=\(\dfrac{8}{15}\)
30 tháng 3 2017

Gọi A = \(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}\)

\(\dfrac{1}{2}\)A = \(\dfrac{1}{2}.\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}\right)\)

\(\dfrac{1}{2}\)A = \(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)

\(\dfrac{1}{2}\)A = \(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

\(\dfrac{1}{2}\)A = \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(\dfrac{1}{2}\)A = \(\dfrac{1}{3}-\dfrac{1}{10}\)

\(\dfrac{1}{2}\)A = \(\dfrac{7}{30}\)

A = \(\dfrac{7}{30}:\dfrac{1}{2}\)

A = \(\dfrac{7}{15}\)

30 tháng 3 2017

\(A=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{55}+\dfrac{1}{66}\)

\(A=2\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\right)\)

\(A=2\left(\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}\right)\)

\(A=2\left(\left(\dfrac{1}{4}-\dfrac{1}{5}\right)+\left(\dfrac{1}{5}-\dfrac{1}{6}\right)+\left(\dfrac{1}{6}-\dfrac{1}{7}\right)+\left(\dfrac{1}{7}-\dfrac{1}{8}\right)+\left(\dfrac{1}{8}-\dfrac{1}{9}\right)+\left(\dfrac{1}{9}-\dfrac{1}{10}\right)+\left(\dfrac{1}{10}-\dfrac{1}{11}\right)+\left(\dfrac{1}{11}-\dfrac{1}{12}\right)\right)\)

\(A=2\left(\dfrac{1}{4}-\dfrac{1}{12}\right)\Rightarrow A=\dfrac{1}{3}\)