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 = \(\frac{2.2012}{\frac{2.2012}{2013}}=2013\)

\(D=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}\right):\left(\frac{2011}{1}+\frac{2010}{2}+...+\frac{1}{2011}\right)\)

\(\Rightarrow D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)

\(\Rightarrow D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}\)

\(\Rightarrow D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}}\)

\(\Rightarrow D\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{2012\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)}\)

\(\Rightarrow D=\frac{1}{2012}\)

câu 1: tính lần lượt là dc

câu 2:

\(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{2450}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}\)

\(=\frac{49}{50}\)

lên lời giải hay ý

\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+...+\frac{1}{20}.\frac{20.21}{2}=1+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}=1+\frac{24.19}{2}=229\)

Tính biểu thức D

\(D=\frac{2.2012}{1+\frac{2}{2.\left(1+2\right)}+\frac{2}{2\left(1+2+3\right)}+\frac{2}{2\left(1+2+3+4\right)}+...+\frac{2}{2\left(1+2+..+2012\right)}}\)

\(=\frac{2.2012}{1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{4050156}}\)

\(=\frac{2.2012}{1+2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{4050156}\right)}\)

\(=\frac{2.2012}{1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2012.2013}\right)}\)

\(=\frac{2.2012}{1+2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2012}-\frac{1}{2013}\right)}\)

\(=\frac{2.2012}{1+2.\left(\frac{1}{2}-\frac{1}{2013}\right)}\)

\(=\frac{2.2012}{1+\frac{2.2011}{2.2013}}\)

\(=\frac{2.2012}{1+\frac{2011}{2013}}\)

\(=\frac{4024}{\frac{4024}{2013}}\)

\(=2013\)

Vậy D=2013

\(D=\frac{2\cdot2012}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2012}}\)

\(D=\frac{2.2012}{1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{2025078}}\)

\(D=\frac{4024}{1+2\cdot\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{4050156}\right)}\)

\(D=\frac{4024}{2\cdot\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2012\cdot2013}\right)}\)

\(D=\frac{4024}{1+2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)}\)

\(D=\frac{4024}{1+2\cdot\left(\frac{1}{2}-\frac{1}{2013}\right)}\)

\(D=\frac{4024}{1+2.\left(\frac{2013-2}{4026}\right)}\)

\(D=\frac{4024}{1+2\cdot\frac{2011}{4026}}\)

\(D=\frac{4024}{1+\frac{2011}{2013}}\)

\(D=\frac{4024}{\frac{4024}{2013}}\)

\(D=\frac{1}{2013}\)

1=1*2/2

1+2=2*3/2

1+2+3=3*4/2

...

1+2+3+...+2012=2012*2013/2

Thay vào là ra.

mh biết làm bài này rùi bn có cần mi2h đang cho bn ko?

\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2012}=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2012.2013}\)

\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)=2\left(1-\frac{1}{2013}\right)=2.\frac{2012}{2013}\)\(\Rightarrow A=\frac{2.2012}{2.2012:2013}=\frac{1}{2013}\)