Vì lớp 6 chưa học hàng đẳng thức nên phải làm thêm bước này:
ta có : a2 - b2 = a2 - ab + ab - b2 = a(a-b) + b( a-b) = (a-b)(a+b)
⇒ a2 - b2 = (a-b)(a+b)
Áp dụng vào biểu thức A ta có :
A=(1−122).(1−132).(1−142)....(1−1302)
A = ( 1 - \(\dfrac{1}{2}\))(1+ \(\dfrac{1}{2}\))(1 - \(\dfrac{1}{3}\)).(1+\(\dfrac{1}{3}\)).....(1-\(\dfrac{1}{30}\))(1+\(\dfrac{1}{30}\))
A = {(1-\(\dfrac{1}{2}\))(1-\(\dfrac{1}{3}\)).(1-\(\dfrac{1}{4}\))........(1-\(\dfrac{1}{30}\))}{(1+\(\dfrac{1}{2}\))(1+\(\dfrac{1}{3}\)).......(1+\(\dfrac{1}{30}\))}
A =( \(\dfrac{1}{2}\).\(\dfrac{2}{3}\).\(\dfrac{3}{4}\).....\(\dfrac{29}{30}\))( \(\dfrac{3}{2}\).\(\dfrac{4}{3}\).\(\dfrac{5}{4}\).........\(\dfrac{31}{30}\))
A = \(\dfrac{2.3.4.5.6......29}{2.3.4.5.6.....29}\) \(\times\) \(\dfrac{1}{30}\) x \(\dfrac{3.4.5.......30}{3.4.5.......30}\) \(\times\) \(\dfrac{31}{2}\)
A =1 \(\times\) \(\dfrac{1}{30}\) \(\times\) \(\dfrac{31}{2}\)
A = \(\dfrac{31}{60}\)
