(a2-a+2012)(a2-a+2014)-3
đặt a2-a+2012=x => a2-a+2014=x+2
ta có x(x+2)-3
= x2+2x-3
=x2+3x-1x-3x
=x(x+3)-1(x+3)
=(x-1)(x+3)
thay x= a2-a+2012
=> (a2-a+2012-1)(a2-a+2012+3)
=(a2-a+2011)(a2-a+2015)
( a2 - a + 2012)( a2 - a + 2014) - 3
Đặt : a2 - a + 2013 = k , ta có :
( k - 1)( k + 1) - 3
= k2 - 22
= ( k + 2)( k - 2)
Thay , a2 - a + 2013 = k , ta có :
( a2 - a + 2015)( x2 - a + 2011)
( a2 - a + 2012)( a2 - a + 2014) - 3
Đặt : a2 - a + 2013 = k , ta có :
( k - 1)( k + 1) - 3
= k2 - 22
= ( k + 2)( k - 2)
Thay , a2 - a + 2013 = k , ta có :
( a2 - a + 2015)( x2 - a + 2011)