\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{380}\)

\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{19\cdot20}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\)

\(=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)

TL

\(\frac{1}{6}\)+\(\frac{1}{12}\)+\(\frac{1}{20}\)+....+\(\frac{1}{380}\)

=>\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+...+\(\frac{1}{19.20}\)

=>\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)+....+\(\frac{1}{19}\)-\(\frac{1}{20}\)

=>\(\frac{1}{2}\)-\(\frac{1}{20}\)

=>\(\frac{9}{20}\)

HT

Gọi ƯCLN(2n+1005;4n+2011)=d(\(d\in\)N*) 

\(\Rightarrow2n+1005⋮d\Rightarrow4n+2010⋮d\Rightarrow4n+2011-4n-2010⋮d\Leftrightarrow1⋮d\Leftrightarrow d=1\)

Vậy ta có đpcm 

gọi d là ƯC(2n+1005,4n+2011)(d\(\in\)N*) 

theo bài ra ta có 

2n+1005\(⋮\)d\(\Rightarrow\)2(2n+1005)\(⋮\)d\(\Rightarrow\)4n+2010\(⋮\)d

4n+2011\(⋮\)d

\(\Rightarrow\)(4n+2011)-(4n+2010)\(⋮\)d

\(\Rightarrow\)4n+2011-4n+2010\(⋮\)d

\(\Rightarrow\)1\(⋮\)d

\(\Rightarrow\)d=1

vậy với mọi n \(\in\)N thì \(\dfrac{2n+1005}{4n+2011}\) là phân số tối giản

TL

bằng nhau

nha bn

HT

sắt nặng hơn

HT

\(A=\dfrac{1.2.3...19}{2.3.4...20}=\dfrac{1}{20}\)

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{20}\right)\)

\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{19}{20}\)

\(=\frac{1\cdot2\cdot3\cdot...\cdot19}{2\cdot3\cdot4\cdot...\cdot20}\)

\(=\frac{1}{20}\)