Vế trái: = 1/2.(1/14+1/15+1/16+...+1/26)

Vế phải: =1/2.(1/(1.2)+1/(3.4)+...+1/(25.26))

Suy ra: (1/14+1/15+...+1/26)= (1/(1.2)+1/(3.4)+...+1/(25.26))

Ta có: (1/(1.2)+1/(3.4)+...+1/(25.26)) = (1-1/2+1/3-1/4+1/5-1/6+....-1/26)

(Các số có mẫu số chẵn là dấu (-) và mẫu số lẻ là dấu (+))

Chuyển các số -1/14+1/15-1/16+...-1/26 từ vế phải sang vế trái ta có:

(1/14+1/15+1/16+...+1/26)+(1/14-1/15+1/16-1/17+...+1/26) = (1-1/2+1/3-1/4+....+1/13)

Suy ra: 2.(1/14+1/18+1/20+...+1/26)= (1-1/2+1/3-1/4+....+1/13)

Suy ra: (1/7+1/8+1/9+...+/1/13)= (1-1/2+1/3-1/4+....+1/13)

Chuyển từ +1/7-1/8+...+1/13 từ vế phải sang trái ta có:

(1/7+1/8+...+1/13)-1/7+1/8-1/9+...-1/13 = (1-1/2+1/3-1/4+1/5-1/6)

Suy ra: 2.(1/8+1/10+1/12) = (1-1/2+1/3-1/4+1/5-1/6)

Suy ra: (1/4+1/5+1/6) = (1-1/2+1/3-1/4+1/5-1/6)

Chuyển: -1/4+1/5-1/6 từ vế phải sang trái ta có:

(1/4+1/5+1/6) +1/4-1/5+1/6 = (1-1/2+1/3)

Suy ra: 2.(1/4+1/6) = (1-1/2+1/3)

Suy ra: (1/2+1/3) =(1-1/2+1/3) Suy ra 2 vế bằng nhau và bằng 5/6.

\(A=\dfrac{3}{1\cdot4}+\dfrac{3}{2\cdot6}+\dfrac{3}{3\cdot8}+...+\dfrac{1}{2012\cdot1342}\\ =\dfrac{3}{1\cdot4}+\dfrac{3}{2\cdot6}+\dfrac{3}{3\cdot8}+...+\dfrac{3}{2012\cdot4026}\\ =\dfrac{6}{2\cdot4}+\dfrac{6}{4\cdot6}+\dfrac{6}{6\cdot8}+...+\dfrac{6}{4024\cdot4026}\\ =3\cdot\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{4024\cdot4026}\right)\\ =3\cdot\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{4024}-\dfrac{1}{4026}\right)\\ =3\cdot\left(\dfrac{1}{2}-\dfrac{1}{4026}\right)\\ =3\cdot\dfrac{1}{2}-3\cdot\dfrac{1}{4026}\\ =1,5-\dfrac{3}{4026}< 1,5\)

Vậy \(A< 1,5\left(đpcm\right)\)

\(\frac{3}{1.4}+\frac{3}{2.6}+\frac{3}{3.8}+...+\frac{1}{2012.1342}\)

\(=\frac{3}{1.4}+\frac{3}{2.6}+\frac{3}{3.8}+...+\frac{3}{2012.4026}\)

\(=\frac{6}{2.4}+\frac{6}{4.6}+\frac{6}{4.8}+...+\frac{6}{4024.4026}\)

\(=3\cdot\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{4024.4026}\right)\)

\(=3\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{4024}-\frac{1}{4026}\right)\)

\(=3\cdot\left(\frac{1}{2}-\frac{1}{4026}\right)\)

\(=3\cdot\frac{1}{2}-3\cdot\frac{1}{4026}\)

\(=1,5-\frac{3}{4026}< 1,5\)