\(\frac{1}{3}+\frac{13}{15}+...+\frac{9997}{9999}\)

\(=1-\frac{2}{3}+1-\frac{2}{15}+...+1-\frac{2}{9999}\)

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

\(=50-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=50-\left(1-\frac{1}{101}\right)\)

Sau bạn tính tiếp là OK rồi 

Hình như đề bài nhầm thì phải: 61/63 chứ?

\(\dfrac{1}{3}+\dfrac{13}{15}+\dfrac{33}{35}+...+\dfrac{9997}{9999}\)

\(=1-\dfrac{2}{3}+1-\dfrac{2}{15}+1-\dfrac{2}{35}+...+1-\dfrac{2}{9999}\)

\(=\left(1+1+1+...+1\right)-\dfrac{2}{3}+\dfrac{2}{15}+...+\dfrac{2}{9999}\)

\(=50-1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

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

\(=\dfrac{4950}{101}\)

Đặt \(A=\)\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{143}\)

\(=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{11.13}\)

\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\)

\(2A=\frac{1}{3}-\frac{1}{13}=\frac{10}{39}\)

\(A=\frac{5}{39}\)

Câu còn lại cx dựa như vậy nhé bn ! 

Chúc bn hc tốt <3

câu c hình như sai đề hả bn

3/15+3/35+3/63+...+3/9603

=3/3.5+3/5.7+3/7.9+...+3/97.99

=3/2.(2/3.5+2/5.7+2/7.9+...+2/97.99)

=3/2.(1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99)=3/2.(1/3-1/99)

=3/2.32/99

=16/33

\(\frac{16}{33}\)

Ta gọi A= 3/15+3/35+3/63+...+3/9603

A=3/3.5+3/5.7+3/7.9+...+3/97.99

2.A=3.2/3.5+3.2/5.7+...+3.2/97.99

2.A=3/3-3/5+3/5-3/7+...+3/97-3/99

2.A=3/3-3/99

2.A=32/33

  A=32/33:2

  A=16/33

3/15+3/35+3/63+.....+3/9603

=3/3.5+3/5.7+3/7.9+.....+3/97.99

=5-3/3.5+7-5/5.7+9-7/7.9+.....+99-97/97.99

=5/3.5-3/3.5+7/5.7-5/5.7+9/7.9-7/7.9+.....+99/97.99-97/97.99

=1/3-1/5+1/5-1/7+1/7-1/9+.....+1/97-1/99

=1/3-1/99

=32/99

câu trả lời của bạn kia k hiểu gì hết