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Ta có ; \(1\dfrac{1}{10}.1\dfrac{1}{11}.....1\dfrac{1}{2011}.1\dfrac{1}{2012}\)
\(=\dfrac{11}{10}.\dfrac{12}{11}.....\dfrac{2012}{2011}.\dfrac{2013}{2012}\)
\(=\dfrac{2013}{10}=201,3\)
1 \(\dfrac{1}{10}\)x 1\(\dfrac{1}{11}\)x 1\(\dfrac{1}{12}\) x...x 1\(\dfrac{1}{2011}\)x 1\(\dfrac{1}{2012}\)
=\(\dfrac{11}{10}\)x \(\dfrac{12}{11}\)x \(\dfrac{13}{12}\)x \(\dfrac{2012}{2011}\)x \(\dfrac{2013}{2012}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3 :
b) Ta có 1+ 2 + 3 +4 + ...+ x =15
Nên \(\frac{x\left(x+1\right)}{2}=15\)
\(x\left(x+1\right)=30\)
=> \(x\left(x+1\right)=5.6\)
=> x = 5
Bài 2:
h; \(\dfrac{2}{3}\)\(x\) + 50% + \(x\) = \(\dfrac{1}{10}\)
\(\dfrac{2}{3}\)\(x\) + \(\dfrac{1}{2}\) + \(x\) = \(\dfrac{1}{10}\)
(\(\dfrac{2}{3}\)\(x\) + \(x\)) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) (\(\dfrac{2}{3}\) + 1) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) = \(\dfrac{1}{10}\) - \(\dfrac{1}{2}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) = \(\dfrac{-2}{5}\)
\(x\) = \(\dfrac{-2}{5}\): \(\dfrac{5}{3}\)
\(x\) = - \(\dfrac{6}{25}\)
Lớp 5 chưa học số âm em nhé.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2\times x-\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}=\frac{3}{11}\)
\(2\times x-2\times\frac{1}{12}+\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{3}{11}\)
\(2\times\left(x-\frac{1}{12}\right)+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{3}{11}\)
\(2\times\left(x-\frac{1}{12}\right)+\left(\frac{1}{3}-\frac{1}{10}\right)=\frac{3}{11}\)
\(2\times\left(x-\frac{1}{12}\right)+\frac{7}{30}=\frac{3}{11}\)
\(2\times\left(x-\frac{1}{12}\right)=\frac{13}{330}\)
\(x-\frac{1}{12}=\frac{13}{660}\)
\(x=\frac{17}{165}\)
\(2x-\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}=\frac{3}{11}\)
\(\Rightarrow2x-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)=\frac{3}{11}\)
\(\Rightarrow2x-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{3}{11}\)
\(\Rightarrow2x-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{3}{11}\)
\(\Rightarrow2x-\left(\frac{1}{2}-\frac{1}{10}\right)=\frac{3}{11}\)
\(\Rightarrow2x-\frac{2}{5}=\frac{3}{11}\)
\(\Rightarrow2x=\frac{3}{11}+\frac{2}{5}\)
\(\Rightarrow2x=\frac{37}{55}\)
\(\Rightarrow x=\frac{37}{55}:2\)
\(\Rightarrow x=\frac{37}{110}\)
Vậy \(x=\frac{37}{110}\)
_Chúc bạn học tốt_
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(1+\dfrac{1}{10}\right)\times\left(1+\dfrac{1}{11}\right)\times...\times\left(1+\dfrac{1}{20}\right)\\ =\dfrac{10+1}{10}\times\dfrac{11+1}{11}\times...\times\dfrac{20+1}{20}\\ =\dfrac{11}{10}\times\dfrac{12}{11}\times\dfrac{13}{12}\times...\times\dfrac{21}{20}\\ =\dfrac{21}{10}\)