\(\frac{1}{n}.\frac{1}{n+4}=\frac{1}{4}.\frac{n+4-n}{n\left(n+4\right)}=\frac{1}{4}.\left(\frac{1}{n}-\frac{1}{n+4}\right)\)
\(\frac{1}{n}.\frac{1}{n+4}=\frac{1}{4}.\frac{n+4-n}{n\left(n+4\right)}=\frac{1}{4}.\left(\frac{1}{n}-\frac{1}{n+4}\right)\)
chứng minh
\(\frac{1}{n}\cdot\frac{1}{n+4}=\frac{1}{4}\cdot\left(\frac{1}{n}-\frac{1}{n+4}\right)\)
Chứng minh rằng
\(\frac{1\cdot3\cdot5\cdot\cdot\cdot\left(2n-1\right)}{\left(n+1\right)\cdot\left(n+2\right)\cdot\left(n+3\right)\cdot...\cdot2n}=\frac{1}{2^n}\)
Chứng minh rằng:
a)\(\frac{1\cdot3\cdot5\cdot\cdot\cdot39}{21\cdot22\cdot23\cdot\cdot\cdot40}=\frac{1}{2^{20}}\)
b)\(\frac{1\cdot3\cdot5\cdot\cdot\cdot\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\cdot\cdot\cdot2n}=\frac{1}{2^n}\)Với \(n\inℕ^∗\)
Chứng minh rằng :
\(\frac{1\cdot3\cdot5\cdot...\cdot\left(2n-1\right)}{\left(n+1\right)\cdot\left(n+2\right)\cdot\left(n+3\right)\cdot...\cdot2n}=\frac{1}{2^n}\)
tính nhanh
a, \(\frac{-2}{5}\cdot\left(\frac{5}{17}-\frac{9}{15}\right)-\frac{2}{5}\cdot\frac{2}{17}+\frac{-2}{5}\)
b, \(\frac{1}{5}\cdot\left(\frac{4}{13}-\frac{9}{11}\right)+\frac{1}{3}\left(\frac{9}{13}-\frac{4}{22}\right)\)
c, \(\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{3}+1\right)\cdot\left(\frac{1}{4}+1\right)\cdot...\cdot\left(\frac{1}{99}+1\right)\)
d, \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{100}\right)\)
Chứng minh:\(\left(1-\frac{2}{6}\right)\left(1-\frac{2}{12}\right)\left(1-\frac{2}{20}\right)....\left(1-\frac{2}{n\left(n+1\right)}\right)>\frac{1}{3}\left(nEN\cdot\right)\)
a) \(\left(\frac{11}{4}\cdot\frac{-5}{9}-\frac{4}{9}\cdot\frac{11}{4}\right)\cdot\frac{8}{33}\)
b) \(\frac{-1}{4}\cdot\frac{152}{11}+\frac{68}{4}\cdot\frac{-1}{11}\)
c) \(\frac{-2}{3}\cdot\frac{4}{5}+\frac{2}{3}\cdot\frac{3}{5}\)
d) \(\left(\frac{1}{2}-1\right)\cdot\left(\frac{1}{3}-1\right)\cdot\left(\frac{1}{4}-1\right)\cdot....\cdot\left(\frac{1}{100}-1\right)\)
e) \(\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot...\cdot\frac{8^{99}}{30^2}\)
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{5}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\cdot\left(1-\frac{1}{2018}\right)\)
Tìm số tự nhiên m,n biết:
\(\frac{1}{4}\cdot\left(m-n\right)\cdot\left(m+n\right)\cdot\left[1+\left(-1\right)^{m+n}\right]=2011\)