ta có A= 6/2.5 + 6/5.8+...+ 6/47.50

             =2x( 3/2x5 +3/5x8 +...+3/47x50)

             =2x(1/2 - 1/5 + 1/5 - 1/8 + ... + 1/47 - 1/50)

             =2x(1/2 - 1/50 ) =1-2/5= 5/5 -2/5 =3/5 

bài này quá dễ 

Giải:

Ta có:

\(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{\left(x-3\right)\times x}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x-3}-\frac{1}{x}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{x}=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}\Leftrightarrow x=3\)

De thoi ma,minh ko ghi lai de nha

=\(\frac{1}{2}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{8}\)+...+...=\(\frac{1}{6}\)

=\(\frac{1}{2}\)-\(\frac{1}{x}\)=\(\frac{1}{6}\)

Con lai bn tu lam nha . chuc bn hok tot !!!

\(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+....+\frac{3}{\left(x-3\right).x}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{\left(x-3\right)}-\frac{1}{x}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{x}=\frac{1}{2}-\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{x}=\frac{1}{3}\)

\(\Leftrightarrow x=1\div\frac{1}{3}=3\)

\(\frac{1}{2\times5}+\frac{1}{5\times8}+\frac{1}{8\times11}+\frac{1}{11\times14}+\frac{1}{14\times17}+\frac{1}{17\times20}\)

\(=\frac{1}{3}\times\left(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+\frac{3}{14\times17}+\frac{3}{17\times20}\right)\)

\(=\frac{1}{3}\times\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)

\(=\frac{1}{3}\times\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(=\frac{1}{3}\times\frac{9}{20}\)

\(=\frac{3}{20}\)

_Chúc bạn học tốt_

Đặt \(A=\frac{1}{2x5}+\frac{1}{5x8}+..+\frac{1}{17x20}\)

\(3xA=3x\left(\frac{1}{2x5}+\frac{1}{5x8}+...+\frac{1}{17x20}\right)\)

\(3xA=\frac{3}{2x5}+\frac{3}{5x8}+....+\frac{3}{17x20}\)

\(3xA=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+..+\frac{1}{17}-\frac{1}{20}\)

\(3xA=\frac{1}{2}-\frac{1}{20}\)

\(3xA=\frac{9}{20}\)

\(\Rightarrow A=\frac{3}{20}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)

\(=\frac{1}{2}-\frac{1}{14}\)

\(=\frac{7}{14}-\frac{1}{14}\)

\(=\frac{6}{14}\)

\(=\frac{3}{7}\)

3/2x5 + 3/5x8 + 3/8x11 + 3/11x14 

= 3/2 - 3/5 + 3/5 - 3/8 + 3/8 - 3/11 + 3/11 - 3/14 

= 3/2 - 3/14 

= 21/14 - 3/14 

= 18/14 

= 9/5 

\(B=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}=\frac{1}{2}-\frac{1}{14}=\frac{3}{7}\)

lên mạng mà tra

\(\frac{9}{2.5}+\frac{39}{5.8}+\frac{87}{8.11}+...+\frac{9897}{98.101}\)

\(=1-\frac{1}{2.5}+1-\frac{1}{5.8}+1-\frac{1}{8.11}+...+1-\frac{1}{98.101}\)

\(=1+1+...+1-\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{98.101}\right)\)              \(\left(\text{33 chữ số 1}\right)\)

\(=33-\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{98.101}\right)\)

\(=33-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\right)\)

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

\(=3-\frac{1}{3}-\frac{99}{202}\)

\(=\frac{1319}{606}\)

\(\frac{2}{2x5}+\frac{2}{5x8}+\frac{2}{8x11}+...+\frac{2}{96x98}=\frac{2}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{96}-\frac{1}{98}\right)\)

                                                                      \(=\frac{2}{3}\left(\frac{1}{2}-\frac{1}{98}\right)=\frac{2}{3}x\frac{24}{49}=\frac{16}{49}\)

= 2 x ( 1/2 x 5 + 1/ 5 x 8 + 1/ 8 x 11 + 1/ 11 x 14 + 1/ 14 x 17 )

= 2 x ( 1/2 - 1/5 + 1/5 - 1/8 + ....+1/14 - 1/17)

= 2 x (1/2 - 1/17)

= 2 x 15/34

= 15/17

ĐÚNG THÌ TÍCH CHO MÌNH NHA

CHÚC BẠN HỌC GIỎI

Đặt \(A=\frac{2}{2.5}+\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+\frac{2}{14.17}\)

\(A=\frac{2}{3}\cdot\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{14}-\frac{1}{17}\right)\)

\(A=\frac{2}{3}\cdot\left(\frac{1}{2}-\frac{1}{17}\right)\)

\(A=\frac{2}{3}\cdot\frac{15}{34}=\frac{5}{17}\Rightarrow A< 1\)

   \(\frac{2}{2.5}+\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+\frac{2}{14.17}\)

\(=\frac{2}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\right)\)

\(=\frac{2}{3}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\right)\)

\(=\frac{2}{3}.\left(1-\frac{1}{17}\right)\)

\(=\frac{2}{3}.\frac{16}{17}\)

\(=\frac{32}{51}< \frac{51}{51}=1\)

\(\Rightarrow\frac{2}{2.5}+\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+\frac{2}{14.17}< 1\)

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