Year 2016

Algebra Level 2

x(x+5)2016(x3)2017(6x)1231(x2)10000(x+1)2015(4x)2420\dfrac{x{(x+5)}^{2016}{(x-3)}^{2017}{(6-x)}^{1231}}{{(x-2)}^{10000}{(x+1)}^{2015}{(4-x)}^{242}} \geq 0

Find all the possible values of xx.

Clarification: In the options, \cup stands for union.

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