[(x^2+2)/(x^2-3x)-(x+1)/(x-3)]/[1/(x-3)+1]
=(x^2+2-x^2-x)/[x(x-3)]/[(1+x-3)/(x-3)]
=-(x-2)/[x(x-3)]/[(x-2)/(x-3)]
=-1/x
∵x=(2/1)^(-1)=1/2
∴原式=-1/x=-2
Y(X) = [(x^2+2)/(x^2-3x)-(x+1)/(x-3)]/[1/(x-3)+1]
=(1+2X+2/X)/(X-2)
X=1/2 代入Y(X)
Y(1/2) = (1+1+4)/(-3/2) = -6/1.5 = -4
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