Bài 2:
+; A = [tex] 1 + \frac{1}{3} + \frac{1}{6}+\frac{1}{10}+ .......... +\frac{2}{x(x+1} = \frac{2011}{1006} [/tex]
0,5 A = [tex] \frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+ .......... +\frac{1}{x(x+1)} = \frac{2011}{2012} [/tex]
0,5 A = [tex] 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3}+ .......... + \frac{1}{x(x+1)} = \frac{2011}{2012} [/tex]
0,5 A = [tex] 1 + \frac{1}{x(x+1)} = \frac{2011}{2012} [/tex]
0,5 A = [tex] \frac{1}{x(x+1)} = \frac{-1}{2012} [/tex] ( bơt cả 2 vê cho 1 )
0,5 A = [tex] \frac{1}{x(x+1)} = \frac{1}{-2012} [/tex]
mà [tex] \frac{1}{x(x+1) \geq 0 [/tex]
=> Không hơp lí
=> Chữa đê đi