compute
(2014²-2020) (2014²+4028-3) (2015)÷(2011)(2013)(2016)(2017)
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3
(2014²-2020)(2014²+4028-3)(2015)÷(2011)(2013)(2016)(2017)
(4056196-2020)(4056196+4028-3)÷[(2011)(2013)][(2016)(2017)]
(4054176)(4060221)÷(4048143)(4054176)
33168613823785440÷16460850532896
= 2015
(4056196-2020)(4056196+4028-3)÷[(2011)(2013)][(2016)(2017)]
(4054176)(4060221)÷(4048143)(4054176)
33168613823785440÷16460850532896
= 2015
Anonymous:
It could be done by writing numeratoras (2014^2 - 2014 - 6)(2014^2 - 2x2014 -3)x2015 .first two brackets are equal to (2011)(2016) and (2013)(2017) which cancel out with the denominator. hence ans = 2015
Answered by
2
calculation is give below
(2014²-2020)(2014²+4028-3)(2015) ÷ (2011)(2013)(2016)(2017)
(4056196-2020)(4056196+4028-3) ÷ [(2011)(2013)][(2016)(2017)]
(4054176) (4060221) ÷ (4048143)(4054176)
33168613823785440÷1646085053289= 2015
plz mark it best
(2014²-2020)(2014²+4028-3)(2015) ÷ (2011)(2013)(2016)(2017)
(4056196-2020)(4056196+4028-3) ÷ [(2011)(2013)][(2016)(2017)]
(4054176) (4060221) ÷ (4048143)(4054176)
33168613823785440÷1646085053289= 2015
plz mark it best
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