\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+....+\frac{1}{9999}\)

\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+....+\frac{1}{99.101}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+....+\frac{1}{99}-\frac{1}{101}\)

\(A=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\) (vì tất cả các phân số khác ngoài 1/3 và 1/101 đều đã bị cộng với số đối với nó = 0)

A= 1/15+1/35+1/63+1/99+……+1/9999
=1/(3×5)+1/(5×7)+1/(7×9)+1/(9×11)+……+1/(99×101)
=1/2(1-1/3)+1/2(1/3-1/5)+1/2(1/5-1/7)+1/2(1/7-1/9)+1/2(1/9-1/11)+……+1/2(1/99-1/101)
=1/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+……+1/99-1/101)
=1/2(1-1/101)
=1/2×(100/101)
=50/101  SAI 

A x 2 = 1 - 1/15 + 1/63 - 1/99 + ... + 1/9999

Ax2= 1-1/9999

Ax2= 9998/9999

A=9998/9999 : 2

A= 4999/9999

A = 1/3.5 + 1/5.7 + 1/7.9 + 1/9.11 + ... + 1/99.101

= 1/2.(2/3.5 + 2/5.7 + 2/7.9 + ... + 2/99.101)

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

= 1/2.(1/3 - 1/101)

= 1/2.98/303

= 49/303

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

cách giai 

chỉ giùm mình đi

A= 1/3 x 5 + 1/5x7 + 1/7x9 + 1/9x11+....+1/99x101

2A= 2/3x5 + 2/5x7 + 2/7x9 + 2/9x11 +....+ 2/99x101

2A= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/11 + ....+ 1/99 - 1/101

2A= 1/3 - 1/101

A= ( 1/3 - 1/101 ) : 2 = 98/303 : 2 = 49/303

A=1/3x5+1/5x7+1/7x9+1/9x11+...+1/99x101

A=1/2x(1/3-1/5+1/5-1/7+1/7-1/9+...+1/99-1/101

A=1/2x(1/3-1/101)

A=1/2x(98/103)

A=49/103

  A = 1/3x5 + 1/5x7 + 1/7x9 + 1/9x11 +...+ 1/99x101

2A = 2/3x5 + 2/5x7 + 2/7x9 + 2/9x11 +...+ 2/99x101

2A = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 +....+ 1/93 - 1/97

2A = 1 - 1/97

2A = 96/97

 A = 96/97 : 2

A = 48/97

 k mk nha bn!!!!^_^

\(A=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{99\times101}\)

\(=\frac{1}{2}\times\left(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{99\times101}\right)\)

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

\(=\frac{1}{2}\times\frac{98}{303}=\frac{49}{303}\)

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