Simplify $$\frac{{{x^2} + 2x + {y^2}}}{{{x^3} - 5{x^2}}}{\text{if }}x + \frac{{{y^2}}}{x} = 5.$$

Solution (By Examveda Team)

$$\eqalign{ & {\text{Given:}} \cr & x + \frac{{{y^2}}}{x} = 5 \cr & {\text{Put }}y = 2,\,x = 1 \cr & {\text{Now,}} \cr & \frac{{{x^2} + 2x + {y^2}}}{{{x^3} - 5{x^2}}} \cr & = \frac{{{{\left( 1 \right)}^2} + 2 \times 1 + {{\left( 2 \right)}^2}}}{{{{\left( 1 \right)}^3} - 5{{\left( 1 \right)}^2}}} \cr & = \frac{7}{{ - 4}} \cr & {\text{Put }}y = 2{\text{ in option and option D will satisfy the condition}} \cr & \cr & {\bf{Alternate \,solution:}} \cr & x + \frac{{{y^2}}}{x} = 5 \cr & {x^2} + {y^2} = 5x \cr & {x^2} - 5x = - {y^2} \cr & {\text{Now, }}\frac{{{x^2} + 2x + {y^2}}}{{{x^3} - 5{x^2}}} \cr & = \frac{{5x + 2x}}{{x\left( {{x^2} - 5x} \right)}} \cr & = \frac{{7x}}{{x\left( { - {y^2}} \right)}} \cr & = - \frac{7}{{{y^2}}} \cr} $$