\(1+\frac{1+\frac{2}{2}}{1+\frac{2}{3}}\)
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\(\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{14}}{-1-\frac{3}{7}+\frac{3}{28}}\)
\(=\frac{\frac{20}{21}-\frac{1}{14}}{\frac{-10}{7}+\frac{3}{28}}\)
\(=\frac{\frac{37}{42}}{\frac{-37}{28}}\)
Đặt \(A=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.......+\frac{2}{2015}+\frac{1}{2016}\)
\(=\frac{2015}{2}+1+\frac{2014}{3}+1+...........+\frac{1}{2015}+1\)
\(=\frac{2017}{2}+\frac{2017}{3}+.........+\frac{2017}{2015}+\frac{2017}{2016}\)
\(=2017.\left(\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{2015}+\frac{1}{2016}\right)\)
Thay A vào biểu thức ta dc
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2017}}{A}\)
\(=\frac{\frac{1}{2017}}{2017}\)\(=1\)
CÓ THỂ LÀ SAI NÊN BẠ THÔNG CẢM CHO MK
\(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+......+\dfrac{1}{1+2+......+50}\)
\(=\dfrac{1}{\dfrac{2.3}{2}}+\dfrac{1}{\dfrac{3.4}{2}}+\dfrac{1}{\dfrac{4.5}{2}}+......+\dfrac{1}{\dfrac{50.51}{2}}\)
\(=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+........+\dfrac{2}{50.51}\)
\(=2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+......+\dfrac{1}{99.100}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)
\(=2.\dfrac{49}{100}\)
\(=\dfrac{49}{50}\)
Xét thừa số tổng quát: \(1+2+3+...+n=\dfrac{n\left(n+1\right)}{2}\)
Suy ra: \(\dfrac{1}{1+2+3+...+n}=\dfrac{1}{\dfrac{n\left(n+1\right)}{2}}=\dfrac{2}{n\left(n+1\right)}\)
Dễ r:v
a)
\(\begin{array}{l}M = \frac{1}{2} + \frac{2}{3} + \left( { - \frac{1}{2}} \right) + \frac{1}{3}\\ = \frac{3}{6} + \frac{4}{6} + \left( {\frac{{ - 3}}{6}} \right) + \frac{2}{6}\\ = \frac{{3 + 4 + \left( { - 3} \right) + 2}}{6}\\ = \frac{6}{6} = 1\end{array}\)
b)
\(\begin{array}{l}M = \frac{1}{2} + \frac{2}{3} + \left( { - \frac{1}{2}} \right) + \frac{1}{3}\\ = \left[ {\frac{1}{2} + \left( {\frac{{ - 1}}{2}} \right)} \right] + \left[ {\frac{2}{3} + \frac{1}{3}} \right]\\ = 0 + 1 = 1\end{array}\)
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
\(1+\frac{1+\frac{2}{2}}{1+\frac{2}{3}}=1+\frac{2}{1+\frac{2}{\frac{5}{3}}}=1+\frac{2}{1+2\cdot\frac{3}{5}}\)
\(=1+\frac{2}{1+\frac{6}{5}}=1+\frac{2}{\frac{11}{5}}=1+2\cdot\frac{5}{11}=1+\frac{10}{11}\)
\(=\frac{21}{11}\)
\(1+\frac{1+\frac{2}{2}}{1+\frac{2}{3}}\)
\(=1+\frac{2}{\frac{5}{3}}\)
\(=1+\frac{6}{5}\)
\(=\frac{11}{5}\)