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

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

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

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

\(\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+...+\frac{6}{19.21}\)

\(=3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)\)

\(=3\left(\frac{1}{3}-\frac{1}{21}\right)\)

\(=3.\frac{2}{7}\)

\(=\frac{6}{7}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

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

\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

~ Hok tốt ~

\(\)

Viết thành 2 . (1/3.5 + 1/5.7 + 1/7.9 + ...+ 1/97.99

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

= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)

\(=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)

bạn làm thiếu rồi

phải chia 2 nữa

Đặt tên bthuc là A

\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{19.21}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{21}\)

\(2A=1-\frac{1}{21}=\frac{20}{21}\)

=>\(A=\frac{20}{21}:2=\frac{10}{21}\)

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{17.19}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{17}-\frac{1}{19}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{19}\right)=\frac{1}{2}.\left(\frac{18}{19}\right)\)

\(=\frac{9}{19}\)

x=9/2

nha bạn

x = 9/2

       \(\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{97\times99}\)

\(=\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{9}\right)+...+\left(\frac{1}{97}-\frac{1}{99}\right)\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

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

\(=\frac{99-3}{297}\)

\(=\frac{96}{297}=\frac{32}{99}\)