Đặt \(A=\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}\)

\(2A=\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}\)

\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\)

\(2A=\frac{1}{1}-\frac{1}{9}=\frac{8}{9}\)

\(A=\frac{8}{9}.\frac{1}{2}=\frac{4}{9}\)

A= 1/(1x3) + 1/(3x5)+ 1/(5x7) + 1/(7x9) +  1/(9x11)

A x 2 = 2/(1x3) + 2/(3x5)+ 2/(5x7) + 2/(7x9) +  2/(9x11)

Nhận xét :

2/(1x3) = 1 - 1/3

2/(3x5) = 1/3 - 1/5

2/(5x7) = 1/5 - 1/7

2/(7x9) = 1/7 - 1/9

2/(9x11) = 1/9 - 1/11

A x 2 = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11

A x 2 = 1 - 1/11

A x 2 = 10/11

A = 10/11 : 2 = 5/11

các bạn k mình nha!

A= 1/(1x3) + 1/(3x5)+ 1/(5x7) + 1/(7x9) +  1/(9x11)

A x 2 = 2/(1x3) + 2/(3x5)+ 2/(5x7) + 2/(7x9) +  2/(9x11)

Nhận xét :

2/(1x3) = 1 - 1/3

2/(3x5) = 1/3 - 1/5

2/(5x7) = 1/5 - 1/7

2/(7x9) = 1/7 - 1/9

2/(9x11) = 1/9 - 1/11

A x 2 = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11

A x 2 = 1 - 1/11

A x 2 = 10/11

A = 10/11 : 2 = 5/11

Coi A=1/1x3+1/3x5+1/5x7+1/7x9

=>2A=2x(1/1x3+1/3x5+1/5x7+1/7x9)=2/1x3+2/3x5+2/5x7+2/7x9

=1-1/3+1/3-1/5+1/5-1/7+1/7-1/9

=1-1/9=8/9

=>A=8/9:2=4/9

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

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

\(=\frac{1}{2}.\frac{100}{101}\)

\(=\frac{50}{101}\)

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{99\cdot101}\)

\(=2\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{99\cdot101}\right)\)

\(=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{99\cdot101}\)

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

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

1/1 x 3 + 1/3 x 5 + 1/5 x 7 + 1/7 x 9 + 1/9 x 11

= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11

= 1 - 1/11

= 10/11

\(\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{9.11}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{1}{3.5}+....+\frac{2}{9.11}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{11}\right)=\frac{1}{2}.\left(1-\frac{1}{11}\right)\)

\(=\frac{1}{2}.\frac{10}{11}=\frac{5}{11}\)

sửa đề câu a  và câu b  nhá  , mik nghĩ đề như này :

  \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)

 \(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)

= \(\frac{1}{1}-\frac{1}{215}\)

\(=\frac{214}{215}\)

b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)

    \(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)

\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)

\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)

\(A\cdot2=\frac{214}{215}\)

\(A=\frac{214}{215}:2\)

\(A=\frac{107}{215}\)

@ミ★Ŧɦươйǥ★彡 cảm ơn bạn nhiều

trả lời hiền thương đề bài của bạn ấy là đúm gòi nha

                            Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

                              \(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

                             \(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

                            \(2A=1-\frac{1}{9.11}=1-\frac{1}{99}=\frac{98}{99}\)

                              \(A=\frac{98}{99}:2=\frac{49}{99}\)

                                Ủng hộ mk nha!!!

A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

A = \(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

A = \(\frac{1}{2}.\left(1-\frac{1}{11}\right)=\frac{1}{2}.\frac{10}{11}\)

A = \(\frac{5}{11}\)