X=1

Nha bạn 

x=1 nha ban

\(\Rightarrow x\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...........+\frac{1}{28}-\frac{1}{33}\right)=\frac{-7}{6}\)

\(\Rightarrow x.\left(\frac{1}{3}-\frac{1}{33}\right)=\frac{-7}{6}\)

\(\Rightarrow x.\frac{10}{33}=\frac{-7}{6}\)

\(\Rightarrow x=\frac{-7}{6}:\frac{10}{33}\)

\(\Rightarrow x=\frac{-231}{60}\)

a)\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{2015}\right)\)

\(=\frac{3}{2}\cdot\frac{402}{2015}\)

\(=\frac{603}{2015}\)

b)\(=\frac{4}{5}\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{98}\right)\)

\(=\frac{4}{5}\left(\frac{1}{3}-\frac{1}{98}\right)\)

\(=\frac{4}{5}\cdot\frac{95}{294}\)

\(=\frac{38}{147}\)

a) Gọi tổng trên là A

A = \(\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+...+\frac{3}{2013.2015}\)

A == \(\frac{3}{5}-\frac{3}{7}+\frac{3}{7}-\frac{3}{9}+\frac{3}{9}-\frac{3}{11}+...+\frac{3}{2013}-\frac{3}{2015}\)

Vì một số trừ cho a rồi cộng cho a sẽ bằng chính số đó nên:

A = \(\frac{3}{5}-\frac{3}{2015}\)

A = \(\frac{1209}{2015}-\frac{3}{2015}\)

A = \(\frac{1206}{2015}\)

b) Gọi tổng trên là B

B = \(\frac{4}{3.8}+\frac{4}{8.13}+\frac{4}{13.15}+...+\frac{4}{93.98}\)

B = \(\frac{4}{3}-\frac{4}{8}+\frac{4}{8}-\frac{4}{13}+\frac{4}{13}-\frac{4}{15}+...+\frac{4}{93}-\frac{4}{98}\)

Vì một số trừ cho a rồi cộng cho a sẽ bằng chính số đó nên:

B = \(\frac{4}{3}-\frac{4}{98}\)

B = \(\frac{686}{294}-\frac{12}{294}\)

B = \(\frac{674}{294}=\frac{337}{147}\)

bai thang dung roi

\(\frac{10}{3.8}+\frac{10}{8.13}+\frac{10}{13.18}+...+\frac{10}{48.53}\)

\(=\frac{10}{5}\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{48}-\frac{1}{53}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{53}\right)\)

\(=2.\frac{50}{159}=\frac{100}{159}\)

\(A=\frac{10}{3.8}+\frac{10}{8.13}+\frac{10}{13.18}+\frac{10}{18.23}+\frac{10}{23.28}\)

\(A=2\left(\frac{5}{3.8}+\frac{5}{8.13}+\frac{5}{13.18}+\frac{5}{18.23}+\frac{5}{23.28}\right)\)

\(A=2\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{23}-\frac{1}{28}\right)\)

\(A=2\left(\frac{1}{3}-\frac{1}{28}\right)\)

\(A=2.\frac{25}{84}=\frac{25}{42}\)

\(A=\frac{10}{3\cdot8}+\frac{10}{8\cdot13}+\frac{10}{13\cdot18}+\frac{10}{18\cdot23}+\frac{10}{23\cdot28}\)

\(A=10\left(\frac{1}{3\cdot8}+\frac{1}{8\cdot13}+\frac{1}{13\cdot18}+\frac{1}{18\cdot23}+\frac{1}{23\cdot28}\right)\)

\(A=\frac{10}{5}\left(\frac{5}{3\cdot8}+\frac{5}{8\cdot13}+\frac{5}{13\cdot18}+\frac{5}{18\cdot23}+\frac{5}{23\cdot28}\right)\)

\(A=2\cdot\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+\frac{1}{23}-\frac{1}{28}\right)\)

\(A=2\cdot\left(\frac{1}{3}-\frac{1}{28}\right)\)

\(A=2\cdot\frac{25}{84}\)

\(A=\frac{25}{42}\)

\(A=\frac{10}{3.8}+\frac{10}{8.13}+\frac{10}{13.18}+\frac{10}{18.23}+\frac{10}{23.28}\)

   \(=2\left(\frac{5}{3.8}+\frac{5}{8.13}+\frac{5}{13.18}+\frac{5}{18.23}+\frac{5}{23.28}\right)\)

   \(=2\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+\frac{1}{23}-\frac{1}{28}\right)\)

   \(=2\left(\frac{1}{3}-\frac{1}{28}\right)\)

    \(=2.\frac{25}{84}\)

    \(=\frac{25}{42}\)

Study well ! >_<

\(B=\frac{10}{3\cdot8}+\frac{10}{8\cdot13}+\frac{10}{13\cdot18}+\frac{10}{18\cdot23}+\frac{10}{23\cdot28}\)

\(B=2\left[\frac{5}{3\cdot8}+\frac{5}{8\cdot13}+\frac{5}{13\cdot18}+\frac{5}{18\cdot23}+\frac{5}{23\cdot28}\right]\)

\(B=2\left[\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{23}-\frac{1}{28}\right]\)

\(B=2\left[\frac{1}{3}-\frac{1}{28}\right]=\frac{25}{42}\)

B = 10/3.8 + 10/8.13 + 10/13.18 + 10/18.23 + 10/23.28

   = 2.( 5/3.8 + 5/8.13 + 5/13.18 + 5/18.23 + 10/23.28 )

   = 2.( 1/3 -1/8 + 1/8 - 1/13 + 1/13 - 1/18 + 1/18 - 1/23 + 1/23 - 1/28 )

   = 2.( 1/3 - 1/28 )

   = 2. 25/84

   = 25/42

\(\frac{24}{42}\) nha bn  chúc hc tốtttt

A = \(\frac{1}{1.4}\)+ \(\frac{1}{4.7}\)+\(\frac{1}{7.10}\)+...+ \(\frac{1}{2014.2017}\)
3A = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{2014.2017}\)
3A = \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{2014}-\frac{1}{2017}\)
3A= 1 - \(\frac{1}{2017}\)
A = \(\frac{1}{3}-\frac{1}{2017.3}\)
A = \(\frac{672}{2017}\)

Ta có \(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2014.2017}\)

\(\Rightarrow A=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2014}-\frac{1}{2017}\right)\)

\(\Rightarrow A=\frac{1}{3}.\left(1-\frac{1}{2017}\right)\)

\(\Rightarrow A=\frac{1}{3}.\frac{2016}{2017}=\frac{672}{2017}\)

Vậy \(A=\frac{672}{2017}\)

~ Học tốt

# Chiyuki Fujito

672/2017