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25 tháng 7 2017

\(\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+....+\frac{2}{x\left(x+1\right)}=\frac{11}{40}\)

\(\Leftrightarrow\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+....+\frac{2}{x\left(x+1\right)}=\frac{11}{40}\)

\(\Leftrightarrow\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+....+\frac{2}{x\left(x+1\right)}=\frac{11}{40}\)

\(\Leftrightarrow2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{11}{40}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+1}=\frac{11}{40}\div2\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{5}-\frac{11}{80}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{16}\Rightarrow x=16-1=15\)

29 tháng 7 2018

A)  7/38 x 9/11 +7/38 x 4/11 -7/38 x 2/11

=7/38.(9/11+4/11-2/11)

=7/38

B) 5/31 x 21/25 + 5/31 x -7/10 - 5/31 x 9/20

=5/31.(21/25-7/10-9/20)

=5/31.(-31/100)

=-1/20

1 tháng 7 2018

Sửa đề : \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Leftrightarrow\)\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\Leftrightarrow\)\(\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}.\frac{1}{2}\)

\(\Leftrightarrow\)\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)

\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{18}\)

\(\Leftrightarrow\)\(x+1=18\)

\(\Leftrightarrow\)\(x=18-1\)

\(\Leftrightarrow\)\(x=17\)

Vậy \(x=17\)

Chúc bạn học tốt ~ 

1 tháng 7 2018

\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)

\(\Leftrightarrow\)\(x+10\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)=\frac{3}{11}\)

\(\Leftrightarrow\)\(x+10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)

\(\Leftrightarrow\)\(x+10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)

\(\Leftrightarrow\)\(x+10.\frac{4}{55}=\frac{3}{11}\)

\(\Leftrightarrow\)\(x+\frac{40}{55}=\frac{3}{11}\)

\(\Leftrightarrow\)\(x=\frac{3}{11}-\frac{40}{55}\)

\(\Leftrightarrow\)\(x=\frac{-5}{11}\)

Vậy \(x=\frac{-5}{11}\)

Chúc bạn học tốt ~ 

24 tháng 8 2018

90 nha 

30 tháng 4 2018

\(\frac{1}{15}+\frac{1}{21}+...+\frac{2}{x.\left(x+1\right)}=\frac{806}{2015}\)

\(\Rightarrow2.\left(\frac{1}{30}+\frac{1}{42}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{806}{2015}\)

\(\Rightarrow2.\left(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{806}{2015}\)

\(\Rightarrow2.\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{806}{2015}\)

\(\Rightarrow2.\left(\frac{1}{5}-\frac{1}{x}\right)=\frac{806}{2015}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x}=\frac{806}{2015}:2\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x}=\frac{403}{2015}\)

\(\Rightarrow\frac{1}{x}=\frac{1}{5}-\frac{403}{2015}\)

\(\Rightarrow\frac{1}{x}=\frac{403}{2015}-\frac{403}{2015}\)

\(\Rightarrow\frac{1}{x}=0\)

\(\Rightarrow x=0\)

Vậy \(x=0\)

Chúc bạn học tốt !!!! 

30 tháng 4 2018

\(\Rightarrow\frac{1}{2}\left(\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{806}{2015}.\frac{1}{2}\)

\(\Rightarrow\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{x\left(x+1\right)}=\frac{403}{2015}\)

\(\Rightarrow\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x\left(x+1\right)}=\frac{403}{2015}\)

\(\Rightarrow\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+...+\frac{x+1-x}{x\left(x+1\right)}=\frac{403}{2015}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{403}{2015}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x+1}=\frac{403}{2015}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{5}-\frac{403}{2015}\)

rồi bạn tự giải nốt nhé

\(A=\dfrac{1}{5}\left(\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{496}-\dfrac{1}{501}\right)\)

\(=\dfrac{1}{5}\cdot\dfrac{55}{334}=\dfrac{11}{334}\)

\(B=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{21}=\dfrac{20}{21}\)

7 tháng 8 2019

\(1,\frac{2}{3}+\frac{4}{9}+\frac{1}{5}+\frac{2}{15}+\frac{3}{2}-\frac{17}{18}\)

\(< =>\frac{4}{9}+\frac{3}{2}+\left(\frac{2}{3}+\frac{1}{5}+\frac{2}{15}\right)-\frac{17}{18}\)

\(< =>\frac{8}{18}+\frac{27}{18}+\left(\frac{10}{15}+\frac{3}{15}+\frac{2}{15}\right)-\frac{17}{18}\)

\(< =>\frac{35}{18}+1-\frac{17}{18}\)

\(< =>\frac{53}{18}-\frac{17}{18}\)

\(< =>2\)

\(2,\frac{13}{28}\cdot\frac{5}{12}-\frac{5}{28}\cdot\frac{1}{12}\)

\(< =>\left(\frac{13}{28}-\frac{5}{28}\right)\cdot\left(\frac{5}{12}-\frac{1}{12}\right)\)

\(< =>\frac{2}{7}\cdot\frac{1}{3}\)

\(< =>\frac{2}{21}\)

\(3,\frac{19}{4}\cdot\frac{15}{23}-\frac{15}{4}\cdot\frac{7}{23}+\frac{15}{4}\cdot\frac{11}{23}\)

\(< =>\frac{285}{92}-\frac{105}{92}+\frac{165}{92}\)

\(< =>\frac{15}{4}\)

8 tháng 8 2019

cảm ơn bạn nha bạn chắc chăn đúng không

18 tháng 7 2018

a) \(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)

\(x-\frac{20}{2}.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)

\(x-10.\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)

\(x-10.\frac{4}{55}=\frac{3}{11}\)

\(x-\frac{8}{11}=\frac{3}{11}\)

x = 1

b) \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\) ( nhân cho cả tử và mẫu của các số hạng trên ( ngoại trừ 2/x.(x+1) ) là 2)

\(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)

\(2.\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(2.\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\frac{1}{x+1}=\frac{1}{18}\)

=> x + 1 = 18

x = 17

18 tháng 7 2018

\(a,x-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)

\(\Rightarrow x-10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)

\(\Rightarrow x-10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)

\(\Rightarrow x-\frac{8}{11}=\frac{3}{11}\)

\(\Rightarrow x=1\)

\(b,\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{18}=\frac{1}{x+1}\)

\(\Rightarrow x+1=18\Leftrightarrow x=17\)