Đặt B = \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(\Rightarrow2B=\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{101-99}{99.101}\)

\(\Rightarrow2B=\frac{5}{3.5}-\frac{3}{3.5}+\frac{7}{5.7}-\frac{5}{5.7}+...+\frac{101}{99.101}-\frac{99}{99.101}\)

\(\Rightarrow2B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(\Rightarrow2B=\frac{1}{3}-\frac{1}{101}\)

\(\Rightarrow2B=\frac{98}{303}\)

\(\Rightarrow B=\frac{98}{303}\div2=\frac{49}{303}\)

bài bạn hỏi tớ có thể giải được nhưng tớ ngại tính lắm

\(\frac{1}{3.5}=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)\)

\(\frac{1}{5.7}=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)\)

\(⋮\)

\(\frac{1}{99.101}=\frac{1}{2}.\left(\frac{1}{99}-\frac{1}{101}\right)\)

=  \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\)

=  \(\frac{1}{2}.\frac{98}{303}\)

=  \(\frac{49}{303}\)

=>  \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}=\frac{49}{303}\)

Mik nghĩ đề phải là cộng chứ

\(=\dfrac{1}{3}-\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{1}{3}-\dfrac{1}{2}\cdot\dfrac{98}{303}=\dfrac{1}{3}-\dfrac{49}{303}=\dfrac{101-49}{303}=\dfrac{52}{303}\)

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{101}\right)\)

\(=\frac{1}{2}.\frac{98}{303}\)

\(=\frac{49}{303}\)

Đặt A=1/1*3+1/3*5+..+1/99*101

A=2/2*(1/1*3+1/3*5+...+1/99*101)

A=1/2*(2/1*3+2/3*5+..+2/99*101)

A=1/2*(1/1-1/3+1/3-1/5+...+1/99-1/100)

A=1/2*(1/1-1/100)

A=1/2*99/100

A=99/200

50/101 nha

Ai chưa có người yêu thì k và kết bạn với mình nhé

\(\frac{1}{1\cdot3}\)+ ... +\(\frac{1}{99\cdot101}\)

2 lần cái này bằng \(\frac{2}{1\cdot3}\)+\(\frac{2}{99\cdot101}\)

= 1/1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/100

=1-1/100

=> cái này bằng 1-1/100 chia 2 = 99/200

Nên nhớ, tao đang học lớp 6 đấy nhé.

A=1/1x3+1/3x5+1/5x7+...+1/99x101

gấp cả 2 vế lên 2 lần ta có:

Ax2=2/1x3+2/3x5+2/5x7+...+2/99x101

Ax2=1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101

Ax2=1-1/101

Ax2=100/101

A=100/101:2=50/101

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}\)

\(=\frac{100}{101}\)

Chúc bạn học tốt nha !!!

1/1.3 + 1/3.5 + 1/5.7 + ... + 1/99.101

= 1/2.(2/1.3 + 2/3.5 + 2/5.7 + ... + 2/99.101)

= 1/2.(1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/99 - 1/101)

= 1/2.(1 - 1/101)

= 1/2.100/101

= 50/101

\(\text{Đặt : }A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{99.101}\)

\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)

\(\Rightarrow2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)

\(\Rightarrow2A=1-\frac{1}{101}\)

\(\Rightarrow A=\frac{100}{101}:2=\frac{50}{101}\)

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(=\frac{1}{1}-\frac{1}{101}=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)

\(\Rightarrow A=\frac{100}{101}:2=\frac{100}{101}\times\frac{1}{2}=\frac{50}{101}\)

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\) 

\(=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\) 

=\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{101}\right)\)

\(=\frac{1}{2}.\frac{98}{303}\)

\(=\frac{49}{303}\)

\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.101}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.......+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\)

\(=\frac{1}{2}.\frac{98}{101}=\frac{49}{101}\)

Hình như Nguyễn Hữu Thế trừ sai.

1/3 - 1/101 = 98/303

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{1}{2}.\dfrac{100}{101}=\dfrac{50}{101}\)