Problem: If x,yx, yx,y and 2x+y22 x+\dfrac{y}{2}2x+2y are not zero, then (2x+y2)−1[(2x)−1+(y2)−1]\left(2 x+\dfrac{y}{2}\right)^{-1}\left[(2 x)^{-1}+\left(\dfrac{y}{2}\right)^{-1}\right](2x+2y)−1[(2x)−1+(2y)−1] equals
Answer Choices:
A. 111
B. xy−1x y^{-1}xy−1
C. x−1yx^{-1} yx−1y
D. (xy)−1(x y)^{-1}(xy)−1
E. none of these Solution:
(2x+y2)−1[(2x)−1+(y2)−1]=12x+y2(12x+1y/2)\left(2 x+\dfrac{y}{2}\right)^{-1}\left[(2 x)^{-1}+\left(\dfrac{y}{2}\right)^{-1}\right]=\dfrac{1}{2 x+\dfrac{y}{2}}\left(\dfrac{1}{2 x}+\dfrac{1}{y / 2}\right)(2x+2y)−1[(2x)−1+(2y)−1]=2x+2y1(2x1+y/21)
=12x+y2(y2+2x)xy=1xy=(xy)−1.=\dfrac{1}{2 x+\dfrac{y}{2}} \dfrac{\left(\dfrac{y}{2}+2 x\right)}{x y}=\dfrac{1}{x y}=(x y)^{-1} . =2x+2y1xy(2y+2x)=xy1=(xy)−1.