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- \(\frac{2.2016}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+..+2016}}\)
Mẫu số = \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2016}\)
\(=1+\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+2016\right).2016:2}\)
\(=1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{1}{2016.2017}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)
\(=2.\left(1-\frac{1}{2017}\right)\)
\(=\frac{2.2016}{2017}\)
Vậy phân số đề bài cho \(=\frac{2.2016}{\frac{2.2016}{2017}}=2.2016.\frac{2017}{2.2016}=2017\)
Mẫu số của A \(=1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2016}\)
\(=\frac{1}{\left(1+0\right).2:2}+\frac{1}{\left(2+1\right).2:2}+\frac{1}{\left(3+1\right).3:2}+\frac{1}{\left(4+1\right).4:2}+...+\frac{1}{\left(2016+1\right).2016:2}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2016.2017}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2016.2017}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)
\(=2.\left(1-\frac{1}{2017}\right)\)
\(=2.\frac{2016}{2017}=2.2016:2017\)
\(A=\left(2.2016\right):\left(2.2016:2017\right)\)
\(A=2.2016:2:2016.2017\)
\(A=2017\)
\(A=\frac{2.2016}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2016}}\)
\(A=\frac{2.2016}{1+\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+..+\frac{1}{2016.2017:2}}\)
\(A=\frac{4032}{1+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2016.2017}}\)
\(A=\frac{4032}{1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{2016.2017}\right)}\) .
\(A=\frac{4032}{1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2016}-\frac{1}{2017}\right)}\)
\(A=\frac{4032}{1+2\left(\frac{1}{2}-\frac{1}{2017}\right)}=\frac{4032}{1+\frac{2015}{2017}}\)
\(A=2017\)
Mẫu số \(=1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2016}\)
\(=\frac{1}{\left(0+1\right).2:2}+\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+\frac{1}{\left(1+4\right).4:2}+...+\frac{1}{\left(1+2016\right).2016:2}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2016.2017}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2016.2017}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)
\(=2.\left(1-\frac{1}{2017}\right)\)
\(=2.\frac{2016}{2017}=2.2016:2017\)
\(A=\left(2.2016\right):\left(2.2016:2017\right)\)
\(A=2.2016:2:2016.2017\)
\(A=2017\)
\(S=\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}+\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\)
\(S=\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{2}{3}-\frac{2}{3}\right)+\left(\frac{3}{4}-\frac{3}{4}\right)-\left(\frac{4}{5}-\frac{4}{5}\right)+\frac{5}{6}\)
\(S=0-0+0-0+\frac{5}{6}\)
\(S=0+\frac{5}{6}\)
\(S=\frac{5}{6}\)
Cẩn thận nha mấy bn, bài này dễ sai dấu lém đó ng kq vẫn đúng
Ủng hộ mk nha ^_^
\(S=\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}+\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\)
\(\Rightarrow S=\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{3}{4}-\frac{3}{4}\right)-\left(\frac{4}{5}-\frac{4}{5}\right)-\left(\frac{2}{3}-\frac{2}{3}\right)+\frac{5}{6}\)
\(\Rightarrow S=\frac{5}{6}\)
a, \(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\left(\frac{2011}{1}+1\right)+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{2012\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)}=\frac{1}{2012}\)
b, \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{2}{2015}+\frac{1}{2016}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}{\left(\frac{2016}{1}+1\right)+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)+1}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{\frac{2017}{1}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{2017\cdot\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}=\frac{1}{2017}\)
Xét mẫu số của F :
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+..+2016}=1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{2016\cdot2017}{2}}\)
\(=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\right)=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{2017}\right)=2-\frac{2}{2017}=\frac{4032}{2017}\)
Suy ra : \(F=\frac{2.2016}{\frac{4032}{2017}}=\frac{2.2016.2017}{4032}=2017\)