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\(C=\frac{2013\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2014}\right)}{\frac{2012}{2}+1+\frac{2011}{3}+1+......+\frac{1}{2013}+1+1}=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{..........1}{2014}\right)}{\frac{2014}{2}+\frac{2014}{3}+.......+\frac{2014}{2013}+\frac{2014}{2014}}\)
\(=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+......+\frac{1}{2014}\right)}{2014.\left(\frac{1}{2}+\frac{1}{3}+......+\frac{1}{2014}\right)}=\frac{2013}{2014}\)
Ta có : 1 + 2 + 3 + ... + n = \(\frac{\left(n+1\right)n}{2}\)
Vậy nên : \(A=2013+\frac{2013}{\frac{3.2}{2}}+\frac{2013}{\frac{4.3}{2}}+...+\frac{2013}{\frac{2013.2012}{2}}\)
\(A=2013+\frac{4026}{2.3}+\frac{4016}{3.4}+...+\frac{4026}{2012.2013}\)
\(A=4026\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}\right)\)
\(A=4026\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)
\(A=4026\left(1-\frac{1}{2013}\right)=4026.\frac{2012}{2013}=4024.\)
Ta có
\(\frac{2014}{1}+\frac{2015}{2}+...+\frac{4026}{2013}=1+1+...+1+\left[\left(\frac{2014}{1}-1\right)+\left(\frac{2015}{2}-1\right)+...+\left(\frac{4026}{2013}-1\right)\right]\)
\(=2013+\left(\frac{2013}{1}+\frac{2013}{2}+...+\frac{2013}{2013}\right)=2013+2013\left(1+\frac{1}{2}+...+\frac{1}{2013}\right)\) (1)
Ta kết hợp (1) và đề
=>\(\left(1+\frac{1}{2}+...+\frac{1}{2013}\right)x+2013=2013+2013\left(1+\frac{1}{2}+...+\frac{1}{2013}\right)\)
=> x=2013
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)x+2013=\frac{2014}{1}+\frac{2015}{2}+...+\frac{4025}{2012}+\frac{4026}{2013}\)
\(\Leftrightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)x=\left(\frac{2014}{1}-1\right)+\left(\frac{2015}{2}-1\right)+...+\left(\frac{4025}{2012}-1\right)+\left(\frac{4026}{2013}-1\right)\)
\(\Leftrightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)x=\frac{2013}{1}+\frac{2013}{2}+...+\frac{2013}{2012}+\frac{2013}{2013}\)
\(\Leftrightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)x=2013\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)\)
\(\Rightarrow x=\frac{2013\left(1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2013}\right)}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}=2013\)
Vậy x = 2013 thoả mãn đề bài.