Hỏi đi hỏi lại vậy.          

1/15+1/35+1/63+1/99+1/143

=1/2x(1/15+1/35+1/63+1/99+1/143)

=1/2x(2/3x5+2/5x7+2/7x9+2/9x11+2/11x13)

=1/2x(1/3-1/5+1/7-1/9+1/9-1/11+1/11-1/13)

=1/2x(1/3-1/13)

=1/2x10/39

=5/39

1/15+1/35+1/63+1/99+1/143

=1/3.5+1/5.7+1/7.9+1/9.11+1/11.13

=1/2.(2/3.5+2/5.7+2/7.9+2/9.11+2/11.13

=1/2.(1/3-1/5+1/5-1/7+1/6-1/9+1/9-1/11+1/11-1/13=1/2.(1/3-1/13)=1/2.10/39=5/39

1/15 + 1/35 + 1/63 + 1/99 + 1/143

đặt A = 1/15 + 1/35 + 1/63 + 1/99 + 1/143

A       = 1/3X5 + 1/5X7 + 1/7X9 + 1/9X11 + 1/11X13

A x 2 = 2 x ( 1/3x5 + 1/5x7 + 1/7x9 + 1/9x11 + 1/11x13 )

A x 2 = 2/3x5 + 2/5x7 + 2/7x9 + 2/9x11 + 2/11x13

A x 2 = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + 1/11 - 1/13

A x 2 = 1/3 - 1/13

A x 2 = 13/39 - 3/39

A x 2 = 10/39

A      =10/39 : 2

A       = 5/39

1/15 + 1/35 + 1/63 + 1/99 + 1/143   = 5/39 nhé bạn !

A=1/3*5+1/5*7+1/7*9+1/9*11+1/11*13

2A=2/3*5+2/5*7+2/7*9+2/9*11+2/11*13=2/3-2/5+2/5-2/7+2/7+2/9-2/9+2/11-2/11+2/13=2/3+2/13=26/39+6/39        =32/39

=>A=16/39

=\(\dfrac{5}{39}\)

\(2A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}=\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{11}-\dfrac{1}{13}\)

\(=\dfrac{1}{3}-\dfrac{1}{13}=\dfrac{13-3}{39}=\dfrac{10}{39}\)

\(\Rightarrow A=\dfrac{10}{39}.\dfrac{1}{2}=\dfrac{10}{78}=\dfrac{5}{39}\)

\(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{25}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)

\(=\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)

\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...+\dfrac{1}{11}-\dfrac{1}{13}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)

\(=\left(1-\dfrac{1}{3}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)

\(=\dfrac{2}{3}\cdot\dfrac{1}{2}+\dfrac{1}{25}\)

\(=\dfrac{1}{3}+\dfrac{1}{25}\)

\(=\dfrac{28}{75}\)

cám ơi bạn

cảm ơn bạn rất nhiều

Đặt phép tính cần tìm là A

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)

\(2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)

\(2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\)

\(2A=1-\dfrac{1}{13}\)

\(2A=\dfrac{12}{13}\)

\(A=\dfrac{6}{13}\)

\(A=\dfrac{1}{3}+\dfrac{1}{15}+...+\dfrac{1}{143}\\ =\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+...+\dfrac{1}{11\times13}\\ =\dfrac{1}{2}\times\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+...+\dfrac{1}{11\times13}\right)\\ =\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{11}-\dfrac{1}{13}\right)\\ =\dfrac{1}{2}\times\dfrac{12}{13}\\ =\dfrac{6}{13}\)

Ta có:\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)

\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{13}\right)\)

\(=\frac{1}{2}.\frac{10}{39}\)

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

B = 1/3*5 + 1/5*7 + 1/7*9 + 1/9*11 + 1/11*13

   = 1/2 * ( 2/3*5 + 2/5*7 + 2/7*9 + 2/9*11 + 2/11*13)

   = 1/2 * ( 1/3 - 1/5 + 1/5 -1/7 + ...+ 1/11 - 1/13)

   = 1/2 * ( 1/3 - 1/11)

   = 1/2 * 8/33

   = 4/33

\(B=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)

\(B=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)

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

\(B=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{13}\right)\)

\(B=\frac{1}{2}.\frac{12}{39}\)

\(B=\frac{2}{13}\)

\(\Rightarrow2B=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}\)

\(\Rightarrow2B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)

\(\Rightarrow\)\(2B=\frac{1}{3}-\frac{1}{13}\)\(=\frac{13}{39}-\frac{3}{39}=\frac{10}{39}\)\(\Rightarrow B=\frac{10}{\frac{39}{2}}=\frac{10}{39\cdot2}=\frac{5}{39}\)

A=1/1.3 + 1/3.5 + 1/5.7 + 1/7.9 + 1/9.11 + 1/11.13 + 1/13.15

A=1/1 - 1/3 +1/3 - 1/5 +1/5 -1/7+......+1/13 - 1/15

A=1 - 1/15

A=1/14

bạn ơi: 

1/195 mới bằng 1/13.15 mà!!

1/3+1/15+1/35+1/63+1/99+1/143+1/180=1/1.3+1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15

=1/2.(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+.....+1/13-1/15)

=1/2.(1-1/15)

=1/2.14/15

=7/15

chúc học tốt