a) \(a=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{37\cdot39}\)

\(a=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)

\(a=\frac{1}{3}-\frac{1}{39}\)

\(a=\frac{12}{39}\)

b) \(\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)\cdot\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)\)

\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)\cdot\left(\frac{-5}{12}+\frac{3}{12}+\frac{2}{12}\right)\)

\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)\cdot\left(\frac{-2}{12}+\frac{2}{12}\right)\)

\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)\cdot0\)

\(=0\)

a) \(A=\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+...+\frac{2}{37x39}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{37}-\frac{1}{39}\)

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

\(A=\frac{4}{13}\)

b) \(\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)x\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)\)

\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)x0\)

\(=0\)

Giải

Ta có A= [1+1/3.5] + [1+1/5.7] + [1+1/7.9] + ... + [1+1/37.39]

=>A= (1+1+1+...+1) +(1/3.5 + 1/5.7 + 1/7.9 + ... + 1/37.39)

=> A = 18 + 1/2.(2/3.5+2/5.7+2/7.9+...+2/37.39)

=>A = 18 + 1/2.(1/3-1/5+1/5-1/7+1/7-1/9+...+1/37-1/39)

=> A= 18 + 1/2.(1/3-1/39)

=> A= 18 + 1/2 . 4/13

=>A= 18 + 2/13 = 236/13

cám ơn bạn 

==>  2A = 2/3x5 + 2/5x7 +....+2/61x63

 2A = 1/3-1/5+1/5-1/7+....+1/61-1/63

 2A = 1/3-1/63

2A=20/63

==> A=10/63

cảm  ơn bạn thần chết.

\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{61.63}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{61}-\frac{1}{63}\)

\(2A=1-\frac{1}{63}\)

\(2A=\frac{62}{63}\)

\(\Rightarrow A=\frac{31}{63}\)

\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.+\frac{1}{101}-\frac{1}{103}\right)\)

\(\frac{1}{2}\left(1-\frac{1}{103}\right)=\frac{1}{2}\cdot\frac{100}{103}=\frac{50}{103}\)

xong r đó

Ta có:

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{101.103}\)

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{101.103}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{103}\right)=\frac{50}{103}\)

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{101\cdot103}\)

\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...+\frac{1}{101}-\frac{1}{103}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{103}\right)\)

\(=\frac{1}{2}\cdot\frac{102}{103}=\frac{102}{206}\)

\(=\frac{51}{103}\)

B=2/3x5 + 2/5x7 + 2/7x9 + ...+2/99x101

B= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 -1/9 + ... + 1/99 - 1/101

B= 1/3 - 1/101

B=98/303

( k mk nhé ! Cách làm câu a và b của mk đều đúng 100% đấy ! Dạng này mk học từ lâu rồi ! )

a, A = 1/2x3+ 1/ 3x4 + 1/4x5 + 1/5x6 + ... + 1/99x100

    A= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 -1/5 + 1/5 - 1/6 + ... + 1/99 -1/100

    A= 1/2 -1/100

    A= 49 / 100

nhờ bạn giải dùm mình câu c,d

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