S = \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{820}\)

S = \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{1640}\)

S = \(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{40.41}\right)\)

S = \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{40}-\frac{1}{41}\right)\)

S = \(2.\left(\frac{1}{2}-\frac{1}{41}\right)\)

S = \(2.\frac{39}{82}\)

S = \(\frac{39}{41}\)

Theo mình nghĩ thì phải nhân S lên 2 nữa chứ.Kết quả là:39/82

Nếu sai thì thông cảm nhé!

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

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

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

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

~ Hok tốt ~

\(\)

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

S = 1/3.5 + 1/5.7 + 1/7.9 +...+1/97.99

S = 1 - ( 1/3+1/3-1/5+1/5-+1/7+1/7+...+1/97-1/99)

S = 1 - 1/99 

S = 98/99

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

\(2S=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{99}\)

\(2S=\frac{1}{3}-\frac{1}{99}\)

\(2S=\frac{33}{99}-\frac{1}{99}\)

\(S=\frac{32}{99}:2\)

\(S=\frac{16}{99}\)

\(S=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\)

\(2S=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)

\(2S=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)

\(2S=\dfrac{1}{3}-\dfrac{1}{99}\)

\(2S=\dfrac{32}{99}\)

\(S=\dfrac{32}{99}:2\)

\(S=\dfrac{16}{99}\)

Cảm ơn bạn rất nhiều! Bạn đã cứu mình rùi! Xin trân thành cảm ơn!

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

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

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

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

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

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

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

Ta có : \(S=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(\Rightarrow2S=2\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\right)\)

\(\Rightarrow2S=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)

\(\Rightarrow2S=\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{99}\right)\)

\(\Rightarrow2S=\left(\frac{1}{3}-\frac{1}{99}\right)+\left[\left(\frac{1}{5}+...+\frac{1}{97}\right)-\left(\frac{1}{5}+\frac{1}{7}+...+\frac{1}{97}\right)\right]\)

\(\Rightarrow2S=\left(\frac{33}{99}-\frac{1}{99}\right)+0\)

\(\Rightarrow2S=\frac{32}{99}\)

\(\Rightarrow S=\frac{32}{99}\div2\)

\(\Rightarrow S=\frac{16}{99}\)

\(S=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\)

\(S=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

\(S=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(S=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(S=\frac{1}{2}.\frac{32}{99}=\frac{16}{99}\)                

2/3.5+ 2 /5.7+ 2/7.9+...+ 2/97.99
=1/3 - 1/5 + 1/5 - 1 /7 +.... + 1/97 - 1/99
=1/3 - 1/99
=32/99

m=/3.5+2/5.7+2/7.9+.....+2/97.99

=m=1/3-1/5+1/5-1/7+.......+1/97-1/99

m=1/3-1/99

=32/99