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Ta có: \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{2012}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2011}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1006}\right)\)
\(=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}\)
\(\Rightarrow A=B\Rightarrow\frac{A}{B}=1\Rightarrow\left(\frac{A}{B}\right)^{2013}=1\)
Vậy \(\left(\frac{A}{B}\right)^{2013}=1\).
S=\(\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)-\left(\frac{1}{2}-\frac{1}{4}-...-\frac{1}{2012}\right)\)
S=\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
S=\(\left(\text{}\text{}\text{}1+\frac{1}{2}+...+\frac{1}{2013}\right)-1-\frac{1}{2}-...-\frac{1}{2012}\)
S=\(\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2013}\)
=>S=P
=>S-P=0
=>(S-P)^2013=0
\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)....\left(1-\frac{1}{x^2}\right)=\frac{1007}{2012}\)
\(\Rightarrow\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{\left(x^2-1\right)}{x^2}=\frac{1007}{2012}\)
\(\Rightarrow\frac{1.3.2.4.3.5...\left(x-1\right)\left(x+1\right)}{\left(2.3.4..x\right)^2}=\frac{1007}{2012}\)
\(\Rightarrow\frac{\left[2.3.4...\left(x-1\right)\right].\left[3.4.5...\left(x+1\right)\right]}{\left(2.3.4....x\right)\left(2.3.4....x\right)}=\frac{1007}{2012}\)
\(\Rightarrow\frac{\left(x+1\right)}{2x}=\frac{1007}{2012}\)
\(\Rightarrow2002.\left(x+1\right)=1007.2x\)
\(\Rightarrow2012x+2012=2014x\)
\(\Rightarrow2x=2012\)
\(\Rightarrow x=1006\)
Vậy x = 1006