Question 173962
{{{(4z)/(8z-12)=(4z)/(4(2z-3))=(cross(4)z)/(cross(4)(2z-3))=z/(2z-3)}}} 


So {{{(4z)/(8z-12)=z/(2z-3)}}} where {{{z<>3/2}}}




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{{{(y^5 - 3y^3 + 8y^2)/(5y)=(y(y^4 - 3y^2 + 8y))/(5y)=(cross(y)(y^4 - 3y^2 + 8y))/(5*cross(y))=(y^4 - 3y^2 + 8y)/5}}}




So {{{(y^5 - 3y^3 + 8y^2)/(5y)=(y^4 - 3y^2 + 8y)/5}}} where {{{y<>0}}}