1/6+1/12+1/20+1/30+...1/x nhân(x+1)=2011/4026
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Đặt \(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(=1-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Rightarrow\)x+1=4026
x=4026-1
x=4025
Vậy x=4025.
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)
=> \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)
=> \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{4026}\)
=> \(1-\frac{1}{x+1}=\frac{2011}{4026}\)
=> \(\frac{1}{x+1}=\frac{2015}{4026}\Rightarrow x+1=\frac{4026}{2015}\Rightarrow x=\frac{2011}{2015}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2013}\)
\(\Rightarrow x+1=2013\)
\(\Rightarrow x=2012\)
Vậy x = 2012
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)\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{x+1}=\frac{9}{10}\)
\(\Leftrightarrow\frac{x+1-1}{x+1}=\frac{9}{10}\)
\(\Rightarrow\frac{x}{x+1}=\frac{9}{10}\)
\(\Rightarrow x=9\)
b) \(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{15-13}{13.15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{15}+x=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{15}\)
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
= \(1-\frac{1}{7}\)
= \(\frac{7}{7}-\frac{1}{7}\)
= \(\frac{6}{7}\)
2) \(\frac{7}{4}-x.\frac{4}{3}=\frac{5}{19}\)
\(x.\frac{4}{3}=\frac{7}{4}-\frac{5}{19}\)
\(x.\frac{4}{3}=\frac{133}{76}-\frac{20}{76}\)
\(x.\frac{4}{3}=\frac{113}{76}\)
\(x=\frac{113}{76}:\frac{4}{3}\)
\(x=\frac{399}{304}\)
VẬY \(x=\frac{399}{304}\)
b) \(\left(x+\frac{3}{4}\right).\frac{5}{7}=\frac{10}{9}\)
\(\left(x+\frac{3}{4}\right)=\frac{10}{9}:\frac{5}{7}\)
\(x+\frac{3}{4}=\frac{14}{9}\)
\(x=\frac{14}{9}-\frac{3}{4}\)
\(x=\frac{29}{36}\)
Vậy \(x=\frac{29}{36}\)
c) \(x.\frac{1}{2}+\frac{3}{2}.x=\frac{4}{5}\)
\(x.\left(\frac{1}{2}+\frac{3}{2}\right)=\frac{4}{5}\)
\(x.2=\frac{4}{5}\)
\(x=\frac{4}{5}:2\)
\(x=\frac{2}{5}\)
Vậy \(x=\frac{2}{5}\)
Chúc bạn học tốt !!!
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)
=> \(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)
=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{4026}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)
=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}=\frac{1}{2013}\)
=> x + 1 = 2013 => x = 2012
Trả lời:
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{x.\left(x+1\right)}=\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{x.\left(x+1\right)}=\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2013}\)
\(\Leftrightarrow x+1=2013\)
\(\Leftrightarrow x=2012\)
Vậy \(x=2012\)