(0.5 By 2 and 2 by 3) to the power of 2013 by (-2 x 3 by 11) to the power of 2012, Another question:[-(xy2)2]3+[-2(-xy2)3]2, (0.5 By 2 and 2 in 3) to the power of 2013 by (-2 by 3 in 11) to the power of 2012, Another question:[-(xy2)2]3+[-2(-xy2)3]2,

(0.5 By 2 and 2 by 3) to the power of 2013 by (-2 x 3 by 11) to the power of 2012, Another question:[-(xy2)2]3+[-2(-xy2)3]2, (0.5 By 2 and 2 in 3) to the power of 2013 by (-2 by 3 in 11) to the power of 2012, Another question:[-(xy2)2]3+[-2(-xy2)3]2,

Suspect question input 2 2/3 should be 3 2/3
(0.5 By 3 and 2 by 3) to the power of 2013 by (-2 x 3 by 11) to the power of 2012
=[(1/2)*(11/3)]^2013*[(-2)*(3/11)]^2012
=[(1/2)*(11/3)]^2012*[(2)*(3/11)]^2012 *(1/2)*(11/3)
=1^2012*11/6
=11/6

2012 Third power +2012 second power –2012 third power –2012 second power –2011

=[2012²(2012-1)-2011]/[2012²(2012+1)-2013]
=2011×(2012²-1)/[2013×(2012²-1)
=2011/2013