\(S=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)

\(S=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.60}\right)\)

\(S=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{60}\right)\)

\(S=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{60}\right)\)

\(S=\frac{3}{2}.\left(\frac{12}{60}-\frac{1}{60}\right)\)

\(S=\frac{3}{2}.\frac{11}{60}\)

\(S=\frac{11}{40}\)

nhi phương k cho mk đi

\(S=\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+...+\frac{3}{59\cdot61}\)

\(S=\frac{3}{2}\left[\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{59\cdot61}\right]\)

\(S=\frac{3}{2}\left[\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{59}-\frac{1}{61}\right]\)

\(S=\frac{3}{2}\left[\frac{1}{5}-\frac{1}{61}\right]=\frac{3}{2}\cdot\frac{56}{305}=\frac{84}{305}\)

Đặt \(A=\frac{3}{5.7}+\frac{3}{7.9}+....+\frac{3}{59.61}\)

 \(\Rightarrow\frac{2}{3}.A=\frac{2}{3}.\left(\frac{3}{5.7}+\frac{3}{7.9}+....+\frac{3}{59.61}\right)\)

\(\Rightarrow\frac{2}{3}.A=\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{59.61}\)

\(\Rightarrow\frac{2}{3}.A=\frac{7-5}{5.7}+\frac{9-7}{7.9}+.....+\frac{61-59}{59.61}\)

\(\Rightarrow\frac{2}{3}.A=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{59}-\frac{1}{61}\)

\(\Rightarrow\frac{2}{3}.A=\frac{1}{5}-\frac{1}{61}\)

\(\Rightarrow\frac{2}{3}.A=\frac{56}{305}\)

\(\Rightarrow A=\frac{56}{305}:\frac{2}{3}\)

\(\Rightarrow A=\frac{56}{305}.\frac{3}{2}\)

\(\Rightarrow A=\frac{84}{305}\)

Vậy \(\frac{3}{5.7}+\frac{3}{7.9}+....+\frac{3}{59.61}=\frac{84}{305}\)

Ta có:

\(A=\frac{6}{5x7}+\frac{6}{7x9}+...\frac{6}{97x99}\)

\(=3x\left(\frac{2}{5x7}+\frac{2}{7x9}+...\frac{2}{97x99}\right)\)

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

\(=3x\left(\frac{1}{5}-\frac{1}{99}\right)\)

\(=3x\left(\frac{99}{495}-\frac{5}{495}\right)\)

\(=3x\frac{94}{495}=\frac{94}{165}\)

Vậy \(A=\frac{94}{165}\)

\(\frac{6}{5}\)x 7 + \(\frac{6}{7}\)x 9 + .... + \(\frac{6}{97}\)x 99

= \(\frac{6}{5}\) - \(\frac{6}{7}\)+\(\frac{6}{7}\)- \(\frac{6}{9}\)+ ..... + \(\frac{6}{97}\)- \(\frac{6}{99}\)

= \(\frac{6}{5}\) - \(\frac{6}{99}\)

= \(\frac{188}{165}\)

nhớ cho đúng đó

\(\frac{3}{5x7}+\frac{3}{7x9}+...+\frac{3}{59x61}\)

\(=\frac{3}{2}\left(\frac{2}{5x7}+\frac{2}{7x9}+...+\frac{2}{59x61}\right)\)

\(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}++...+\frac{1}{59}-\frac{1}{61}\right)\)

\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{61}\right)\)

\(=\frac{3}{2}.\frac{56}{305}=\frac{84}{305}\)

Nguyễn Tuấn Minh giải đúng rồi nhé

Ta có :\(\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{59.61}=2.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)

\(=2\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)=2\left(\frac{1}{5}-\frac{1}{61}\right)=2.\frac{56}{305}=\frac{112}{305}\)

P/S : Dấu "." là dấu "x"

Bài làm:

Ta có: \(\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{59.61}\)

\(=2\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)

\(=2\left(\frac{1}{5}-\frac{1}{61}\right)\)

\(=2.\frac{56}{305}=\frac{112}{305}\)

\(\frac{4}{5\times7}+\frac{4}{7\times9}+.....+\frac{4}{59\times61}\)

\(=\frac{4}{2}\times\left(\frac{2}{5\times7}+\frac{2}{7\times9}+....+\frac{2}{59\times61}\right)\)

\(=2\times\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.....+\frac{1}{59}-\frac{1}{61}\right)\)

\(=2\times\left(\frac{1}{5}-\frac{1}{61}\right)\)

\(=2\times\frac{56}{305}\)

\(=\frac{112}{305}\)

Học tốt

\(\dfrac{6}{3\cdot5}+\dfrac{6}{5\cdot7}+\dfrac{6}{7\cdot9}+.....+\dfrac{6}{33\cdot35}\)

\(=\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{33\cdot35}\right)\cdot3\)

\(=\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+.....+\dfrac{1}{33}-\dfrac{1}{35}\right)\cdot3\)

\(=\left(\dfrac{1}{3}-\dfrac{1}{35}\right)\cdot3\)

\(=\dfrac{32}{3\cdot35}\cdot3\)

\(=\dfrac{32}{35}\)

b)

S2=6/2x5+6/5x8+6/8x11+...+6/29x32

=2.(3/2.5+3/5.8+...+3/29.32)

=2.(1/2-1/5+1/5-1/8+...+1/29-1/32)

=2.(1/2-1/32)

=2.15/32

=15/16

a)

Ta có:

S1=2/3x5+2/5x7+2/7x9+...+2/97x99

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

=1/3-1/99

=32/99

b)

S2=6/2x5+6/5x8+6/8x11+...+6/29x32

=2.(3/2.5+3/5.8+...+3/29.32)

=2.(1/2-1/5+1/5-1/8+...+1/29-1/32)

=2.(1/2-1/32)

=2.15/32

=15/16

 Đúng 1 

a)

Ta có:

S1=2/3x5+2/5x7+2/7x9+...+2/97x99

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

=1/3-1/99

=32/99

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

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

\(=0\)