1/15 +1/35 +1/63 + 1/99 +...+1/2915 +1/3135

=1/3x5+1/5x7+1/7x9+....+1/53x55+1/55x57

=1/3-1/5+1/5-1/7+1/7-1/9+.....+1/53-1/55+1/55-1/57

=1/3-1/57

=6/19 nhé

Ta có:1/15+1/35+1/63+1/99+...+1/2915+1/3135=1/3*5+1/5*7+1/7*9+1/9*11+...+1/53*55+1/55*57

=1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+...+1/53-1/55+1/55-1/57

=1/3-1/57=19/57-1/57=18/57

\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+..+\frac{1}{3135}\)\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+..+\frac{1}{55.57}\)
\(\Rightarrow\)\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\)\(+...+\frac{1}{55}-\frac{1}{57}\)\(=\frac{1}{3}-\frac{1}{57}=\frac{6}{19}\)

 

Đăt S=1/15+1/35+1/63+1/99+...+1/2915+1/3135

         =1/3.5+1/5.7+1/7.9+1/9.11+...+1/53.55+1/55.57

         =1/2(2/3.5+2/5.7+2/7.9+...+2/53.55+2/55.57)

         =1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/53-1/55+1/55-1/57)

         =1/2(1/3-1/57)

         =1/2(19/57-1/57)

         =1/2.18/57

         =3/19

Vậy 1/15+1/35+1/63+1/99+...+1/2915+1/3135=3/19

Mik viết thế này mong bạn thông cảm nha!!

chúc bạn hok tốt!!

Bạn nhớ k cho mik một cái đúng nha!!

Đặt \(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{2915}+\frac{1}{3135}\)

\(\Leftrightarrow A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+....+\frac{1}{53\cdot55}+\frac{1}{55\cdot57}\)

\(\Leftrightarrow2A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{53\cdot55}+\frac{2}{55\cdot57}\)

\(\Leftrightarrow2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-....+\frac{1}{53}-\frac{1}{55}+\frac{1}{55}-\frac{1}{57}\)

\(\Leftrightarrow2A=\frac{1}{3}-\frac{1}{57}=\frac{6}{19}\)

\(\Leftrightarrow A=\frac{6}{19}:2=\frac{3}{19}\)

1/15 + 1/35 + 1/63 + 1/99 + ...+ 1/2915 + 1/3135

= 1/2.(2/15 + 2/35 + 2/63 + 2/99 + ... + 2/2915 + 2/3135)

=1/2.( 2/3.5 + 2/5.7 + 2/7.9 + 2/9.11 + ... + 2/53.55 + 2/55.57)

=1/2.(1/3-1/5+1/5-1/7+1/7-1/9+...+1/53-1/55+1/55-1/57)

=1/2.(1/3-1/57)

=1/2.6/19

=3/19

\(A=\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+.......\frac{1}{13x15}=\frac{1}{2}x\frac{2}{1x3}+\frac{2}{3x5}.......+\frac{2}{13x15}\)

\(A=\frac{1}{2}x\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}\right)\)

Còn lại em nhân giống ở trên nhé

Đặt A = 1/15 + 1/35 + ... + 1/3135 

       A = 1/3.5 + 1/5.7 + ... + 1/55.57

     2A =  2/3.5 + 2/5.7 + ... + 2/55.57 

    2A = 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/55 - 1/57 

    2A = 1/3 - 1/57 = 6/19 

      A = 3/19 

thanks các bn

\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{55\cdot57}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{55}-\frac{1}{57}\)

\(2A=\frac{1}{3}-\frac{1}{57}\)

\(2A=\frac{6}{19}\)

\(A=\frac{3}{19}\)

Ta có:

A=1/15+1/35+1/63+1/99+...+1/2015+1/3135

   =1/3.5+1/5.7+1/7.9+1/9.11+...+1/45.47+1/47.49

   =1/3-1/5+1/5-1/7+1/7-1/9+...+1/45-1/47+1/47-1/49

   =1/3-1/49

   =49/147-3/147

   =47/147

\(A=\frac{1}{12}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{2015}+\frac{1}{3135}\)

\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+...+\frac{1}{45\cdot47}+\frac{1}{47\cdot49}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{47}-\frac{1}{49}\)

\(A=\frac{1}{3}-\frac{1}{49}\)

\(A=\frac{46}{147}\)

                             \(49-3=46\text{ nha bạn !}\)

a) \(\frac{1}{3}+\frac{5}{6}:\left(x-2\frac{1}{5}\right)=\frac{3}{4}\)

=> \(\frac{1}{3}+\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{3}{4}\)

=> \(\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{3}{4}-\frac{1}{3}\)

=> \(\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{5}{12}\)

=> \(x-\frac{11}{5}=\frac{5}{6}:\frac{5}{12}\)

=> \(x-\frac{11}{5}=2\)

=> \(x=2+\frac{11}{5}\)

=> \(x=\frac{21}{5}\)

thanks bn

1/15 +1/35 + 1/63 + ... + 1/3135

= 1/3*5 + 1/5*7 + 1/7*9 + ... + 1/55*57

= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/55 - 1/57

= 1/3 - 1/57

= 18/57

1/3 + 1/6 + 1/10 +  ... + 1/x(x + 1)*2 = 2009/2011

=> 2/6 + 2/12 + 2/20 + ... + 2/x(x + 1)*2 = 2009/2011

=> 2(1/6 + 1/12 + 1/20 + ... + 1/x(x + 1)) = 2009/2011

=> 2(1/2*3 + 1/3*4 + 1/4*5 + ... + 1/x(x + 1)) = 2009/2011

=> 2(1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x + 1)  = 2009/2011

=> 2(1/2 - 1/x + 1) = 2009/2011

=> 1/2 - 1/x + 1 = 2009/4022

=> 1/x + 1 = 1/2011

=> x + 1 = 2011

=> x = 2010

vậy-

bấm vào chữ 0 đúng sẽ ra đáp án 

A = 98/303

Cơn Mưa Tình Yêu

A = 1/15 + 1/35 + 1/ 63 + 1/99 + ...+ 1/9999

A     = 1/(3x5) + 1/(5x7) + 1/(7x9) + 1/(9x11) + ... + 1/(99 x 101)

Ax2 = 2/(3x5) + 2/(5x7) + 2/(7x9) + 2/(9x11) + ... + 2/(99 x 101)

Ax2 = 1/3 – 1/5 + 1/5 – 1/7 + 1/7 – 1/9 + 1/9 – 1/11 + ...+ 1/99 – 1/101

Ax2 = 1/3 – 1/101 = 98/303

A    =  98/303 : 2

A    =  49/303

bai toan nay don

49/303

duyệt đi

\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\)

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

\(\Rightarrow\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)

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

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

\(\Rightarrow\frac{1}{2}.\frac{98}{303}\)

\(\Rightarrow\frac{49}{303}\)

1/15 + 1/35 + 1/63 + 1/99 + ... + 1/9999 = 

= 1/(3x5) + 1/(5x7) + 1/(7x9) + ... + 1/(99x101)

= (1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ...+ 1/99 - 1/101) : 2

= (1/3 - 1/101) : 2 

= 98/303 : 2

= 49/303

\(y=\frac{1}{x^2+\sqrt{x}}\)

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