\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right).....\left(1+\frac{1}{99}\right)\)

\(=\left(\frac{2}{2}+\frac{1}{2}\right)\left(\frac{3}{3}+\frac{1}{3}\right)\left(\frac{4}{4}+\frac{1}{4}\right).....\left(\frac{99}{99}+\frac{1}{99}\right)\)

\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)

\(=\frac{3.4.5....100}{2.3.4....99}=\frac{100}{2}=50\)

\(=4\left(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{53.55}\right)\)

\(=4\left(\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-\frac{1}{7}+...+\frac{1}{53}-\frac{1}{55}\right)\)

\(=4\left(\frac{1}{5}-\frac{1}{55}\right)\)

\(=4.\frac{2}{11}\)

\(=\frac{8}{11}\)

\(j,\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{53.55}=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{53}-\dfrac{1}{55}=\dfrac{1}{5}-\dfrac{1}{55}=\dfrac{11}{55}-\dfrac{1}{55}=\dfrac{10}{55}=\dfrac{2}{11}\\ k,\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{99}{100}=\dfrac{1}{100}.\dfrac{2}{2}.\dfrac{3}{3}...\dfrac{99}{99}=\dfrac{1}{100}.1.1...1=\dfrac{1}{100}\)

=1-(1/3.5+1/3.7+1//7.9+...+1/55.57)

=1-1/2.(2/3.5+2/5.7+2/7.9+...+2/55.57)

=1-1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/55-1/57)

=1-1/2(1/3-1/57)

=1-1/2.18/57

=1-9/57

=48/57

=

1-(1/3.5+1/5.7+1/7.9+....+1/53.55+1/55.57)

=1-1/2.[1/3-1/5+1/5-1/7+1/7-1/9+...+1/53-1/55+1/55-1/57]

=1-1/2.[1/3-1/57]

=1-1/2.54/171

=1-28/171

=143/171.

Ta có: 

\(\dfrac{2}{5.7}=\dfrac{7-5}{5.7}=\dfrac{1}{5}-\dfrac{1}{7}\)

\(\dfrac{2}{7.9}=\dfrac{9-7}{7.9}=\dfrac{1}{7}-\dfrac{1}{9}\)

..........

\(\dfrac{2}{53.55}=\dfrac{55-53}{53.55}=\dfrac{1}{53}-\dfrac{1}{55}\)

\(\Rightarrow\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{53.55}=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{5}+...+\dfrac{1}{53}-\dfrac{1}{55}=\dfrac{1}{5}-\dfrac{1}{55}=\dfrac{10}{55}=\dfrac{2}{11}\)

\(=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{53}-\dfrac{1}{55}=\dfrac{1}{5}-\dfrac{1}{55}=\dfrac{2}{11}\)

= 3/2.( 1/5.7 + 1/7.9 +.....+ 1/53.55 )

= 3/2.( 1/5 - 1/7 + 1/7 - 1/9 +......+ 1/53 - 1/55 )

= 3/2.( 1/5 - 1/55 )

= 3/2.( 11/55 - 1/55 )

= 3/2. 10/55

= 3/2 . 2/11

= 3/11

tk mk nha

bài này mk biết , chỉ cần bạn lấy 3 ra và thêm 2 vào

Chào bạn, bạn hãy theo dõi bài giải của mình nhé!

Ta có : 

\(\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}+...+\frac{4}{53.55}\)

\(=\frac{4}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{53.55}\right)\)

\(=2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{53}-\frac{1}{55}\right)\)

\(=2.\left(\frac{1}{5}-\frac{1}{55}\right)=2.\left(\frac{11}{55}-\frac{1}{55}\right)=2.\frac{10}{55}=2.\frac{2}{11}=\frac{4}{11}\)

Có gì không hiểu bạn hỏi lại mình nhé! Chúc bạn học tốt!

Ta có: \(\frac{4}{5.7}+\frac{4}{7.9}+.....+\frac{4}{53.55}\)

Đặt C = \(\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{53.55}\)

     \(\frac{1}{2}C=\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{9}\right)+....+\left(\frac{1}{53}-\frac{1}{55}\right)\)

      \(\frac{1}{2}C=\frac{1}{5}-\frac{1}{55}\)

      \(\frac{1}{2}C=\frac{2}{11}\)

\(C=\frac{2}{11}:\frac{1}{2}\)

Vậy C = \(\frac{4}{11}\)
Có gì sai thì mong bạn thông cảm

Ta có : 

\(A=2.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+............+\frac{2}{53.55}\right)\)

\(\Rightarrow A=2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+..............+\frac{1}{53}-\frac{1}{55}\right)\)

\(\Rightarrow A=2.\left(\frac{1}{5}-\frac{1}{55}\right)=2.\frac{2}{11}=\frac{4}{11}\)

k nha bạn !!!

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