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NT
0
23 tháng 4 2016
Đặt A = 1/2 + 5/6 + 11/12 + 19/20 + ... + 89/90
A = ( 1 - 1/2 ) + ( 1 - 1/12 ) + ( 1 - 1/20 ) + ... + ( 1 - 1/90 )
A = ( 1 + 1 + 1 + ... + 1 + 1 ) - ( 1/2 + 1/6 + 1/20 + ... + 1/90
A = 9 - ( 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/9.10 )
A = 9 - ( 1 - 1/10 )
A = 9 - 9/10
A = 81/10
16 tháng 8 2018
a)\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+x=\frac{3}{5}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+x=\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+x=\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{10}+x=\frac{3}{5}\)
\(\Rightarrow\frac{2}{5}+x=\frac{3}{5}\)
\(\Rightarrow x=\frac{3}{5}-\frac{2}{5}=\frac{1}{5}\)
b)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+x=\frac{1}{3}\)
\(\Rightarrow\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{13}-\frac{2}{15}+x=\frac{1}{3}\)
\(\Rightarrow\frac{2}{3}-\frac{2}{15}+x=\frac{1}{3}\)
\(\Rightarrow\frac{8}{15}+x=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{3}-\frac{8}{15}=-\frac{1}{5}\)
c)\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{x+1}=\frac{9}{10}\)
\(\Leftrightarrow\frac{x+1-1}{x+1}=\frac{9}{10}\)
\(\Rightarrow\frac{x}{x+1}=\frac{9}{10}\)
\(\Rightarrow x=9\)
DD
Đoàn Đức Hà
Giáo viên
10 tháng 1 2021
b) \(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{15-13}{13.15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{15}+x=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{15}\)
GH
0
20 tháng 8 2021
Ta có: \(\dfrac{1}{2}+\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+\dfrac{41}{42}+\dfrac{55}{56}+\dfrac{71}{72}+\dfrac{89}{90}\)
\(=8-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)
\(=8-\left(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=8-1+\dfrac{1}{10}\)
\(=\dfrac{71}{10}\)
NT
1
16 tháng 3 2017
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+.....+\frac{71}{72}+\frac{89}{90}\)
\(=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+....+\left(1-\frac{1}{72}\right)+\left(1-\frac{1}{90}\right)\)
\(=\left(1+1+1+....+1\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)=9-\frac{9}{10}=\frac{81}{10}\)
PH
3
\(2\times x-\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}=\frac{3}{11}\)
\(2\times x-2\times\frac{1}{12}+\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{3}{11}\)
\(2\times\left(x-\frac{1}{12}\right)+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{3}{11}\)
\(2\times\left(x-\frac{1}{12}\right)+\left(\frac{1}{3}-\frac{1}{10}\right)=\frac{3}{11}\)
\(2\times\left(x-\frac{1}{12}\right)+\frac{7}{30}=\frac{3}{11}\)
\(2\times\left(x-\frac{1}{12}\right)=\frac{13}{330}\)
\(x-\frac{1}{12}=\frac{13}{660}\)
\(x=\frac{17}{165}\)
\(2x-\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}=\frac{3}{11}\)
\(\Rightarrow2x-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)=\frac{3}{11}\)
\(\Rightarrow2x-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{3}{11}\)
\(\Rightarrow2x-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{3}{11}\)
\(\Rightarrow2x-\left(\frac{1}{2}-\frac{1}{10}\right)=\frac{3}{11}\)
\(\Rightarrow2x-\frac{2}{5}=\frac{3}{11}\)
\(\Rightarrow2x=\frac{3}{11}+\frac{2}{5}\)
\(\Rightarrow2x=\frac{37}{55}\)
\(\Rightarrow x=\frac{37}{55}:2\)
\(\Rightarrow x=\frac{37}{110}\)
Vậy \(x=\frac{37}{110}\)
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