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\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)x=\frac{2013}{1}+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}\)
\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)x=\left(\frac{2012}{2}+1\right)+...+\left(\frac{2}{2012}+1\right)+\left(\frac{1}{2013}+1\right)+1\)
\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)x=\frac{2014}{2}+...+\frac{2014}{2012}+\frac{2014}{2013}+\frac{2014}{2014}\)
\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)x=2014.\left(\frac{1}{2}+...+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\right)\)
\(x=\frac{2014.\left(\frac{1}{2}+...+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)
\(x=2014\)
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
a, ĐK: \(x+1\ge0\Leftrightarrow x\ge-1\)
Ta có: |3-2x|=x+1
=>\(\orbr{\begin{cases}3-2x=x+1\\3-2x=-x-1\end{cases}\Rightarrow\orbr{\begin{cases}x+2x=3-1\\-x+2x=3+1\end{cases}\Rightarrow}\orbr{\begin{cases}3x=2\\x=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\left(tmđk\right)\\x=4\left(tmđk\right)\end{cases}}}\)
Vậy x=2/3 hoặc x=4
b, Xét VP ta có: \(\frac{2013}{1}+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}=2013+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}\)
\(=1+\left(1+\frac{2012}{2}\right)+\left(1+\frac{2011}{3}\right)+...+\left(1+\frac{2}{2012}\right)+\left(1+\frac{1}{2013}\right)\)
\(=\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2012}+\frac{2014}{2013}+1\)
\(=\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}+\frac{2014}{2014}=2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)\)
=>\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)x=2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)\)
=>\(x=\frac{2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}=2014\)
Vậy x=2014
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Ta có:
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{2012}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}-1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-...-\frac{1}{1006}\)
\(=\frac{1}{1007}+\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2012}+\frac{1}{2013}\left(1\right)\)
Mà \(P=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\left(2\right)\)
Từ (1) và (2)\(\Rightarrow S=P\Rightarrow\left(S-P\right)^{2013}=0^{2013}=0\)
Vậy...
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