Thực hiện phép tính một cách hợp lí:
\(\frac{2.2012}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+2012}}\)
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JA
27 tháng 7 2016
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
13 tháng 11 2016
\(D=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}\right):\left(\frac{2011}{1}+\frac{2010}{2}+...+\frac{1}{2011}\right)\)
\(\Rightarrow D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
\(\Rightarrow D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}\)
\(\Rightarrow D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}}\)
\(\Rightarrow D\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{2012\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)}\)
\(\Rightarrow D=\frac{1}{2012}\)
6 tháng 4 2016
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+...+\frac{1}{20}.\frac{20.21}{2}=1+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}=1+\frac{24.19}{2}=229\)
3 tháng 11 2016
câu 2:
\(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{2450}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
NL
2
S
24 tháng 7 2018
Đặt A là tên biểu thức
Xét mẫu số, ta có: \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+2012}\)
\(=1+\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+3\right).3}{2}}+...+\frac{1}{\frac{\left(1+2012\right).2012}{2}}\)
\(=\frac{2}{2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2012.2013}\)\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)=2\left(1-\frac{1}{2013}\right)=2\cdot\frac{2012}{2013}\)
\(\Rightarrow A=\frac{2.2012}{2\cdot\frac{2012}{2013}}=\frac{2012.2013}{2012}=2013\)
2 tháng 9 2021
\(\frac{2.2012}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+2012}}\)
\(=\frac{2.2012}{1+\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+3\right).3}{2}}+...+\frac{1}{\frac{\left(1+2012\right).2012}{2}}}\)
\(=\frac{2.2012}{\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2012.2013}}\)
\(=\frac{2.2012}{2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)}\)
\(=\frac{2.2012}{2.\left(1-\frac{1}{2013}\right)}=\frac{2.2012}{2.\frac{2012}{2013}}=\frac{2012}{\frac{2012}{2013}}=\frac{2012.2013}{2012}=2013\)
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 = \(\frac{2.2012}{\frac{2.2012}{2013}}=2013\)