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 )

ta có (a-1)² ≥ 0 ∀a

<=> a²-2a+1 ≥ 0

<=>a²+4a-2a+1 ≥ 4a (cộng cả 2 vế va 4a)

<=> a²+2a+1 ≥ 4a

<=> (a+1)² ≥ 4a

CM tương tự ta đc

(b+1)² ≥ 4b

(c+1)² ≥ 4c

Nhân các vế với nhau ta có

[(a+1)²+(b+1)² +(c+1)² ]² ≥ 4a.4b.4c

<=> [(a+1)²+(b+1)² +(c+1)² ]² ≥64abc

<=> [(a+1)²+(b+1)² +(c+1)² ]² ≥64 (vì abc =1)

<=> (a+1)²+(b+1)² +(c+1)² ≥8 (đpcm)

Đặt A = 1/1.3 + 1/3.5 + 1/5.7 +........+ 1/(2n - 1)(2n + 1)
2.A = 2/1.3 + 2/3.5 + 2/5.7 +........+ 2/(2n - 1)(2n + 1)
2.A = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/(2n - 1) - 1/(2n + 1)
2.A = 1 - 1/(2n + 1) = 2n/(2n + 1)
Vậy A = n/(2n + 1)

Đặt A = 1/1.3 + 1/3.5 + 1/5.7 +........+ 1/(2n - 1)(2n + 1)
2.A = 2/1.3 + 2/3.5 + 2/5.7 +........+ 2/(2n - 1)(2n + 1)
2.A = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/(2n - 1) - 1/(2n + 1)
2.A = 1 - 1/(2n + 1) = 2n/(2n + 1)
Vậy A = n/(2n + 1)

tớ làm câu b thôi, câu a nhân 1/2 lên là đc 

\(A=\frac{1}{2}.\left[\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2n-1\right).\left(2n+1\right)}\right)\right]\)

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

\(A=\frac{1}{2}.\left(1-\frac{1}{2n+1}\right)=\frac{1}{2}-\frac{1}{2.\left(2n+1\right)}< \frac{1}{2}\)

p/s: lưu ý không có dấu "=" đâu nhé vì \(\frac{1}{2.\left(2n+1\right)}>0\left(n\text{ thuộc }N\right)\)

\(S=\frac{1.3}{3.5}+\frac{2.4}{5.7}+\frac{3.5}{7.9}+...+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}+...+\frac{1002.1004}{2005.2007}\)

\(\Rightarrow S=\frac{\left(2-1\right)\left(2+1\right)}{\left(2.2-1\right)\left(2.2+1\right)}+\frac{\left(3-1\right)\left(3+1\right)}{\left(3.2-1\right)\left(3.2+1\right)}+...+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}\)

\(+..+\frac{\left(1003-1\right)\left(1003+1\right)}{\left(1003.2-1\right)\left(1003.2+1\right)}\)

\(\Rightarrow S=\frac{1}{4}-\frac{3}{8}\left(\frac{1}{2.2-1}-\frac{1}{2.2+1}\right)+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{3.2-1}-\frac{1}{3.2+1}\right)+...\)

\(+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)+...+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{1003.2-1}-\frac{1}{1003.2+1}\right)\)

\(\Rightarrow S=1002.\frac{1}{4}-1002.\frac{3}{8}\left(\frac{1}{2.2-1}-\frac{1}{2.2+1}+\frac{1}{3.2-1}-...-\frac{1}{1003.2+1}\right)\)

\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2005}-\frac{1}{2007}\right)\)

\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}\left(\frac{1}{3}-\frac{1}{2007}\right)\)

\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}.\frac{668}{2007}\)

\(\Rightarrow S=\frac{501}{2}-\frac{27889}{223}\)

\(\Rightarrow S=125,4372197\)

\(\)

thx  you

tự làm là hạnh phúc của mỗi công dân.

\(=\lim\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)

\(=\lim\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)

\(=\lim\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)=\dfrac{1}{2}\)