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Ta có: \(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2013}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2011}+\dfrac{1}{2013}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1006}\right)\)
\(=\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\)
\(\Rightarrow P-S=\left(\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)-\left(\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2013}\right)=0\)
\(\Rightarrow\left(P-S\right)^{2013}=0^{2013}=0\)
Vậy \(\left(P-S\right)^{2013}=0\)
Ta có :
\(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+..........+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2013}\)
\(=\left(1+\dfrac{1}{3}+..........+\dfrac{1}{2013}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+......+\dfrac{1}{2012}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}\right)-2\left(\dfrac{1}{2}+......+\dfrac{1}{2012}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}\right)-\left(1+\dfrac{1}{2}+......+\dfrac{1}{1006}\right)\)
\(=\dfrac{1}{1007}+\dfrac{1}{1008}+..........+\dfrac{1}{2013}\)
\(\Leftrightarrow S-P=\left(\dfrac{1}{1007}+\dfrac{1}{1008}+....+\dfrac{1}{2013}\right)-\left(\dfrac{1}{1007}+\dfrac{1}{1008}+....+\dfrac{1}{2013}\right)\)
\(\Leftrightarrow S-P=0\)
\(\Leftrightarrow\left(S-P\right)^{2013}=0^{2013}=0\)
\(1+\dfrac{1}{2}+...+\dfrac{1}{2012}+\dfrac{1}{2013}-2\times\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2010}+\dfrac{1}{2012}\right)\)
\(\Rightarrow1+\dfrac{1}{2}+...+\dfrac{1}{2012}+\dfrac{1}{2013}-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1005}+\dfrac{1}{1006}\right)\)
\(\Rightarrow\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\)
\(\Rightarrow S=P\Rightarrow S-P=0\Rightarrow\left(S-P\right)^{2013}=1\)
\(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2013}\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+....+\dfrac{1}{2013}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2012}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2013}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{2012}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{1006}\right)\)
\(=\dfrac{1}{1007}+\dfrac{1}{1008}+....+\dfrac{1}{2013}=P\)
Vậy \(S=P\)
B=\(\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\)
=\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1006}+\dfrac{1}{1007}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)\)- \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1006}\right)\)
=\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1006}+\dfrac{1}{1007}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)\)-2\(\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)\)
=1-\(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...-\dfrac{1}{2012}+\dfrac{1}{2013}\)=S
( A-B)2013 =0
Chúc ban học tốt nhé...!
\(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2013}\\ =\left(1+\dfrac{1}{3}+...+\dfrac{1}{2011}+\dfrac{1}{2013}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)\\ =\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)\\ =\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1006}\right)\\ =\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\\ \Rightarrow S-P=0\\ \Rightarrow\left(S-P\right)^{2018}=0\)
\(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=\dfrac{2013}{1}+\dfrac{2012}{2}+...+\dfrac{2}{2012}+\dfrac{1}{2013}\)
\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=\left(1+\dfrac{2012}{2}\right)+\left(1+\dfrac{2011}{3}\right)+...+\left(1+\dfrac{2}{2012}\right)+\left(1+\dfrac{1}{2013}\right)+1\)
\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=\dfrac{2014}{2}+\dfrac{2014}{3}+...+\dfrac{2014}{2012}+\dfrac{2014}{2013}+\dfrac{2014}{2014}\)
\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=2014.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}\right)\)
\(\Leftrightarrow x=\dfrac{2014.\left(\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2014}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}}\)
\(\Leftrightarrow x=2014\)
Vậy \(x=2014\)
\(VP=\dfrac{2013}{1}+\dfrac{2012}{2}+...+\dfrac{1}{2013}\\ =\dfrac{2012}{2}+1+\dfrac{2011}{3}+1+...+\dfrac{1}{2013}+1+1\\ =\dfrac{2014}{2}+\dfrac{2014}{3}+...+\dfrac{2014}{2013}+\dfrac{2014}{2014}\\ =2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)\)
\(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)\\ x=2014\)
Vậy x = 2014
\(\frac{x-1}{2014}+\frac{x-2}{2013}-\frac{x-3}{2012}=\frac{x-4}{2011}\)
\(\frac{x-1}{2014}+\frac{x-2}{2013}-\frac{x-3}{2012}-\frac{x-4}{2011}=0\)
\(\left(\frac{x-1}{2014}-1\right)+\left(\frac{x-2}{2013}-1\right)-\left(\frac{x-3}{2012}-1\right)-\left(\frac{x-4}{2011}-1\right)=0\)
\(\frac{x-2015}{2014}+\frac{x-2015}{2013}-\frac{x-2015}{2012}-\frac{x-2015}{2011}=0\)
\(\left(x-2015\right).\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)=0\)
Vì \(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\ne0\)
\(\Rightarrow x-2015=0\)
\(x=0+2015\)
\(x=2015\)
\(\dfrac{x+4}{2011}+\dfrac{x+3}{2012}=\dfrac{x+2}{2013}+\dfrac{x+1}{2014}\)
\(\Rightarrow\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1=\dfrac{x+2}{2013}+1+\dfrac{x+1}{2014}+1\)
\(\Rightarrow\dfrac{x+2015}{2011}+\dfrac{x+2015}{2012}-\dfrac{x+2015}{2013}-\dfrac{x+2015}{2014}=0\)
\(\Rightarrow\left(x+2015\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2013}-\dfrac{1}{2014}\right)=0\)
Mà \(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2013}-\dfrac{1}{2014}\ne0\)
\(\Rightarrow x+2015=0\Rightarrow x=-2015\)
Vậy x = -2015
\(\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}+\dfrac{x+1}{2014}\)
\(\Rightarrow\dfrac{x+4}{2011}+\dfrac{x+3}{2012}-\dfrac{x+2}{2013}-\dfrac{x+1}{2014}=0\)
\(\Rightarrow\)\(\left(\dfrac{x+4}{2011}+1\right)+\left(\dfrac{x+3}{2012}+1\right)-\left(\dfrac{x+2}{2013}+1\right)-\left(\dfrac{x+1}{2014}+1\right)=0\)\(\Rightarrow\dfrac{x+2015}{2011}+\dfrac{x+2015}{2012}-\dfrac{x+2015}{2013}-\dfrac{x+2015}{2014}=0\)
\(\Rightarrow\left(x+2015\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2013}-\dfrac{1}{2014}\right)=0\)
\(\Rightarrow x+2015=0\Rightarrow x=-2015\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{2012+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\left(\frac{2012}{2}+1\right)+\left(\frac{2011}{3}+1\right)+...+\left(\frac{1}{2013}+1\right)}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)}\)
\(=\frac{1}{2014}\)
Vậy \(A=\frac{1}{2014}\)
Đặt B=\(2012+\dfrac{2012}{2}+\dfrac{2011}{3}+...+\dfrac{1}{2013}\)
=>B=\(\left(1+\dfrac{2012}{2}\right)+\left(1+\dfrac{2011}{3}\right)+...+\left(1+\dfrac{1}{2013}\right)\)
=\(\dfrac{2014}{2}+\dfrac{2014}{3}+...+\dfrac{2014}{2013}\)
=\(2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right)\)
=>A=\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2013}}{2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right)}=\dfrac{1}{2014}\)
Vậy ...