Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Mẫu số = \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+2012}\)
\(=1+\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+2012\right).2012:2}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2012.2013}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)
\(=2.\left(1-\frac{1}{2013}\right)=\frac{2.2012}{2013}\)
Phân số đề bài cho = \(\frac{2.2012}{\frac{2.2012}{2013}}=2013\)
\(D=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}\right):\left(\frac{2011}{1}+\frac{2010}{2}+...+\frac{1}{2011}\right)\)
\(\Rightarrow D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
\(\Rightarrow D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}\)
\(\Rightarrow D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}}\)
\(\Rightarrow D\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{2012\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)}\)
\(\Rightarrow D=\frac{1}{2012}\)
\(1-\frac{1}{2}+2-\frac{2}{3}+3-\frac{3}{4}+4-\frac{1}{4}-3-\frac{1}{3}-2-\frac{1}{2}-1=\)
\(\left(1+2+3+4-3-2-1\right)-\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\right)=4-3=1\)
\(D=\frac{2.2012}{1+\frac{2}{2.\left(1+2\right)}+\frac{2}{2\left(1+2+3\right)}+\frac{2}{2\left(1+2+3+4\right)}+...+\frac{2}{2\left(1+2+..+2012\right)}}\)
\(=\frac{2.2012}{1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{4050156}}\)
\(=\frac{2.2012}{1+2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{4050156}\right)}\)
\(=\frac{2.2012}{1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2012.2013}\right)}\)
\(=\frac{2.2012}{1+2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2012}-\frac{1}{2013}\right)}\)
\(=\frac{2.2012}{1+2.\left(\frac{1}{2}-\frac{1}{2013}\right)}\)
\(=\frac{2.2012}{1+\frac{2.2011}{2.2013}}\)
\(=\frac{2.2012}{1+\frac{2011}{2013}}\)
\(=\frac{4024}{\frac{4024}{2013}}\)
\(=2013\)
Vậy D=2013
mh biết làm bài này rùi bn có cần mi2h đang cho bn ko?
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2012}=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2012.2013}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)=2\left(1-\frac{1}{2013}\right)=2.\frac{2012}{2013}\)\(\Rightarrow A=\frac{2.2012}{2.2012:2013}=\frac{1}{2013}\)
Co quy luat nay ne em: 1+2=3=2.3:2; 1+2+3=6=3.4:2;...;1+2+3+...+2012=2012.2013:2
Suy ra ta co:
Mau so cua D=1 + 1/(2.3:2) + 1/(3.4:2) + 1/(4.5:2) + .... + 1/(2012.2013:2)
=1 + 2/2.3 + 2/3.4 + 2/4.5 + .... + 2/2012.2013
= 2.[1/2 + 1/2.3 + 1/3.4 + 1/4.5 + .... + 1/2012.2013]
=2.[1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ..... + 1/2012.2013]
=2.[1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +....+1/2012 - 1/2013
=2[1 - 1/2013]
=2.2012/2013
Vay D= 2.2012 / (2.2012:2013)=2013