Question

[Trần Minh Châu]

 giúp mình với

Answer 1

[Nguyễn Lê Phước Thịnh]

Bài 5:

a: \(A=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{99\times100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

b: \(B=\dfrac{1}{1\times3}+\dfrac{2}{3\times7}+\dfrac{1}{7\times9}+\dfrac{3}{9\times15}+\dfrac{6}{15\times27}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\times3}+\dfrac{4}{3\times7}+\dfrac{2}{7\times9}+\dfrac{6}{9\times15}+\dfrac{12}{15\times27}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{27}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{27}\right)=\dfrac{1}{2}\times\dfrac{26}{27}=\dfrac{13}{27}\)

c: \(C=\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+...+\dfrac{1}{2011\times2013}\)

\(=\dfrac{1}{2}\times\left(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+...+\dfrac{2}{2011\times2013}\right)\)

\(=\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2011}-\dfrac{1}{2013}\right)\)

\(=\dfrac{1}{2}\times\left(1-\dfrac{1}{2013}\right)=\dfrac{1}{2}\times\dfrac{2012}{2013}=\dfrac{1006}{2013}\)

Bài 3:

a: Khi m=2006; n=2007; p=2008 thì

\(S=2006\times2+2007\times2+2008\times2=2\times\left(2006+2007+2008\right)=12042\)

b: \(S=m\times2+n\times2+p\times2=2\times\left(m+n+p\right)=2\times2009=4018\)