Question

tôi không biết gõ mấy cía phân số này nên đăng ảnh 

giúp tôi

Answer 1

\(P=\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\)

\(=\left(1+\dfrac{1}{2}+...+\dfrac{1}{1006}+\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)\)

\(=\left(1+\dfrac{1}{2}+...+\dfrac{1}{1006}\right)\)

\(=\left(1+\dfrac{1}{2}+...+\dfrac{1}{1006}+\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)\)

\(=2.\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{2012}\right)\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...-\dfrac{1}{2012}+\dfrac{1}{2013}=S\)

Do đó \(\left(S-P\right)^{2013}=0\)