\(A=-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{25.27}\right)-\frac{1}{27}\)

\(=-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{25}-\frac{1}{27}\right)-\frac{1}{27}\)

\(=-\left(1-\frac{1}{27}\right)-\frac{1}{27}\)

\(=-1+\frac{1}{27}-\frac{1}{27}\)

\(=-1\)

1/2015

ok, cảm ơn bạn

\(A=-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{25.27}\right)-\frac{1}{27}\)

   \(=-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{25}-\frac{1}{27}\right)-\frac{1}{27}\)

    \(=-\left(1-\frac{1}{27}\right)-\frac{1}{27}\)

    \(=-1+\frac{1}{27}-\frac{1}{27}\)

     \(=-1\)

sao làm như vậy được??? 2/19.21 và 2/23.25 bn k làm được như thế đâu vì nó k cùng quy luật với các ps kia

-1

Là -1.Đúng 100%

\(A=-\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{25.27}\right)-\dfrac{1}{27}\)

\(=-\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{25}-\dfrac{1}{27}\right)-\dfrac{1}{27}\)

\(=\)\(-\left(1-\dfrac{1}{27}\right)-\dfrac{1}{27}\)

\(=-1+\dfrac{1}{27}-\dfrac{1}{27}\)

\(=-1+\left(\dfrac{1}{27}-\dfrac{1}{27}\right)\)

\(=-1+0\)

\(=-1\)

a) Đặt B= 1/1.3 + 1/3.5 + 1/5.7 + .....+ 1/19.21

Ta có: 2B= 2/1.3 + 2/3.5 + 2/5.7 + ....+ 2/19.21

= 1- 1/3 + 1/3-1/5 + 1/5-1/7 +....+ 1/19-1/21

= 1-1/21 = 20/21

=> B= 20/21 : 2 => B= 10/21

b) Như trên, ta có: 2A= 1- (1/2n + 1) => A=( 1-1/2n+1).1/2

=> A= 1/2- 1/2n+1

=> A< 1/2 ( đpcm )

Bài này lớp 6 học rùi! 

S = 312/25

Bạn có cần giải cặn kẽ ko

\(2B=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{23}-\frac{1}{25}\right)\)

\(2B=2\left(\frac{1}{3}-\frac{1}{25}\right)\)

\(2B=2\times\frac{22}{75}\)

\(B=\frac{44}{75}\)