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22 tháng 7 2018

\(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{2013}\)

\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{2013}\)

\(\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{2013}\)

\(1-\frac{1}{x+1}=2013\)

\(\frac{x}{x+1}=2013\)

x = 2013x + 2013

Vậy ko có gt của x

11 tháng 12 2017

\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)x=\frac{2013}{1}+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}\)

\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)x=\left(\frac{2012}{2}+1\right)+...+\left(\frac{2}{2012}+1\right)+\left(\frac{1}{2013}+1\right)+1\)

\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)x=\frac{2014}{2}+...+\frac{2014}{2012}+\frac{2014}{2013}+\frac{2014}{2014}\)

\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)x=2014.\left(\frac{1}{2}+...+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\right)\)

\(x=\frac{2014.\left(\frac{1}{2}+...+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)

\(x=2014\)

15 tháng 4 2017

Ta có: \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)x=2013+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}\)

\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)x=1+\left(1+\frac{2012}{2}\right)+...+\left(1+\frac{2}{2012}\right)+\left(1+\frac{1}{2013}\right)\)

\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)x=\frac{2014}{2014}+\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2012}+\frac{2014}{2013}\)

\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)x=2014.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}\right)\)

\(\Rightarrow x=2014\)

Lưu ý: số 2013 ở dòng T2 được tách ra làm 2013 số 1

6 tháng 2 2020

\(a,\left(\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|\right):10=\left(1-\frac{1}{2}\right)....\left(1-\frac{1}{10}\right)\)

\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\Leftrightarrow\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|=1\)

\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.|x-2|=1\Leftrightarrow|x-2|.\frac{2}{3}=1\Leftrightarrow|x-2|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)

6 tháng 2 2020

\(\left(\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|\right):10=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{9}\right).\left(1-\frac{1}{10}\right)\)

\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\)

\(\Leftrightarrow\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|=1\)

\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.\left|x-2\right|=1\)

\(\Leftrightarrow\left|x-2\right|.\frac{2}{3}=1\Leftrightarrow\left|x-2\right|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)