Question

give answer of 28 with calculation

Answer 1

Hi ,

Let 2014 = a ---( 1 )

i ) 2014² - 2020

= 2014² - 2014 - 6

= a² - a - 6

= a² - 3a + 2a - 6

= a( a - 3 ) + 2( a - 3)

= ( a - 3 ) ( a + 2 )

= ( 2014 - 3 ) ( 2014 + 2 )

= 2011 × 2016 ---( 2 )

ii ) 2014² + 4028 - 3

= 2014² + 2 × 2014 - 3

= a² + 2a - 3

= a² + 3a - a - 3

= a( a + 3 ) - 1( a + 3 )

= ( a + 3 )( a - 1 )

= ( 2014 + 3 )( 2014 - 1 )

= 2017 × 2013 ---( 3 )

Now ,

[(2014²-2020)(2014²+4028-3)2015]/[(2011)(2013)(2014)(2017)]

[ From ( 2 ) and ( 3 ) ]

= [(2011)(2016)(2017)(2013)(2015)]/[(2011)(2013)(2016)(2017)]

After cancellation ,

= 2015

I hope this helps you.

: )