Bài 1:

a: \(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\cdots+\frac{2}{97\cdot99}\)

\(=\frac13-\frac15+\frac15-\frac17+\cdots+\frac{1}{97}-\frac{1}{99}\)

\(=\frac13-\frac{1}{99}=\frac{32}{99}\)

b: \(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\cdots+\frac{1}{97\cdot99}\)

\(=\frac12\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\cdots+\frac{2}{97\cdot99}\right)\)

\(=\frac12\left(\frac13-\frac15+\frac15-\frac17+\cdots+\frac{1}{97}-\frac{1}{99}\right)\)

\(=\frac12\left(\frac13-\frac{1}{99}\right)=\frac12\cdot\frac{32}{99}=\frac{16}{99}\)

c: \(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+\cdots+\frac{1}{990}\)

\(=\frac{1}{3\cdot6}+\frac{1}{6\cdot9}+\frac{1}{9\cdot12}+\cdots+\frac{1}{30\cdot33}\)

\(=\frac13\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\cdots+\frac{3}{30\cdot33}\right)\)

\(=\frac13\left(\frac13-\frac16+\frac16-\frac19+\cdots+\frac{1}{30}-\frac{1}{33}\right)\)

\(=\frac13\left(\frac13-\frac{1}{33}\right)=\frac13\cdot\frac{10}{33}=\frac{10}{99}\)

Bài 2:

Sửa đề: \(\frac{1}{41}+\frac{1}{42}+\cdots+\frac{1}{80}>\frac{7}{12}\)

Đặt \(A=\frac{1}{41}+\frac{1}{42}+\cdots+\frac{1}{80}\)

Ta có: \(\frac{1}{41}>\frac{1}{60}\)

\(\frac{1}{42}>\frac{1}{60}\)

...

\(\frac{1}{59}>\frac{1}{60}\)

\(\frac{1}{60}=\frac{1}{60}\)

DO đó: \(\frac{1}{41}+\frac{1}{42}+\cdots+\frac{1}{59}+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+\cdots+\frac{1}{60}+\frac{1}{60}=\frac{20}{60}=\frac13\) (1)

Ta có: \(\frac{1}{61}>\frac{1}{80}\)

\(\frac{1}{62}>\frac{1}{80}\)

...

\(\frac{1}{79}>\frac{1}{80}\)

\(\frac{1}{80}=\frac{1}{80}\)

Do đó: \(\frac{1}{61}+\frac{1}{62}+\cdots+\frac{1}{80}>\frac{1}{80}+\frac{1}{80}+\cdots+\frac{1}{80}=\frac{20}{80}=\frac14\) (2)

Từ (1),(2) suy ra \(\frac{1}{41}+\frac{1}{42}+\cdots+\frac{1}{80}>\frac13+\frac14\)

=>\(A>\frac13+\frac14\)

=>A>7/12

\(=1-\frac{62}{195}\)

\(=\frac{133}{195}\)

Ta có: \(1-\dfrac{2}{3\cdot5}-\dfrac{2}{5\cdot7}-\dfrac{2}{7\cdot9}-...-\dfrac{2}{61\cdot63}-\dfrac{2}{63\cdot65}\)

\(=1-\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{61\cdot63}+\dfrac{2}{63\cdot65}\right)\)

\(=1-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{61}-\dfrac{1}{63}+\dfrac{1}{63}-\dfrac{1}{65}\right)\)

\(=1-\left(\dfrac{1}{3}-\dfrac{1}{65}\right)\)

\(=1-\dfrac{62}{195}\)

\(=\dfrac{133}{195}\)

\(A=\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{37\cdot39}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{37}-\dfrac{1}{39}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{38}{39}< \dfrac{1}{2}\)

\(B=\frac{2^3}{3.5}+\frac{2^3}{5.7}+....+\frac{2^3}{101.103}\)

\(\Rightarrow\frac{1}{2^2}.B=\frac{2}{3.5}+\frac{2}{4.7}+....+\frac{2}{101.103}\)

\(\Rightarrow\frac{1}{4}.B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\)

\(\Rightarrow\frac{1}{4}.B=\frac{1}{3}-\frac{1}{103}=\frac{100}{309}\)

\(\Rightarrow B=\frac{100}{309}:\frac{1}{4}=\frac{400}{309}\)

\(=2^2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{101.103}\right)\)

\(=4\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{103}\right)\)

\(=4\left(\frac{1}{3}-\frac{1}{103}\right)\)

\(=4\cdot\frac{100}{309}=\frac{400}{309}\)

\(1-\dfrac{2}{3.5}-\dfrac{2}{5.7}-...-\dfrac{2}{61.63}-\dfrac{2}{63.65}\)

\(=1-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{63}-\dfrac{1}{65}\right)\)

\(=1-\left(\dfrac{1}{3}-\dfrac{1}{65}\right)\)

\(=1-\dfrac{62}{195}\)

\(=\dfrac{133}{195}\)

=1-(1/3-1/5+1/5-1/7+...+1/61-1/63)

=1-20/63=43/63

Ta có: \(B=1-\dfrac{2}{3.5}-\dfrac{2}{5.7}-\dfrac{2}{7.9}-...-\dfrac{2}{61.63}-\dfrac{2}{63.65}\)

\(=1-\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{61.63}+\dfrac{2}{63.65}\right)\)

\(=1-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{63}-\dfrac{1}{65}\right)\)

\(=1-\left(\dfrac{1}{3}-\dfrac{1}{65}\right)\)

\(=1-\dfrac{62}{195}=\dfrac{133}{195}.\)

Vậy \(B=\dfrac{133}{195}.\)

Thank bạn nhiều nha!

Ta có: \(A=\frac{2}{1\cdot3}-\frac{4}{3\cdot5}+\frac{6}{5\cdot7}-\frac{8}{7\cdot9}-\cdots-\frac{20}{19\cdot21}\)

\(=\frac12\left(\frac{4}{1\cdot3}-\frac{8}{3\cdot5}+\frac{12}{5\cdot7}-\frac{16}{7\cdot9}-\cdots-\frac{40}{19\cdot21}\right)\)

\(=\frac12\left(1+\frac13-\frac13-\frac15+\cdots-\frac{1}{19}-\frac{1}{21}\right)\)

\(=\frac12\left(1-\frac{1}{21}\right)=\frac12\cdot\frac{20}{21}=\frac{10}{21}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+.................+\frac{2}{97.99}\)

=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+..................+\frac{1}{97}-\frac{1}{99}\)

=\(\frac{1}{3}-\frac{1}{99}\)

=\(\frac{32}{99}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}\)

\(=\frac{32}{99}\)