2. For real numbers $x$ and $y$, where $x \neq 0, y \notin\{-2,0,2\}$ and $x+y \neq 0$, simplify the expression:

$$
\frac{x y^{2018}+2 x y^{2017}}{y^{2016}-4 y^{2014}} \cdot\left(\left(\frac{x^{2}}{y^{3}}+x^{-1}\right):\left(x y^{-2}-\frac{1}{y}+x^{-1}\right)\right): \frac{(x-y)^{2}+4 x y}{1+\frac{y}{x}}-\frac{y^{2}+2 y}{y+2}
$$

Time for solving: 45 minutes.
2. The numerator of the first fraction $x y^{2018}+2 x y^{2017}$ is transformed into $x y^{2017}(y+2)$, and the denominator $y^{2016}-$ $4 y^{2014}$ is transformed into $y^{2014}(y-2)(y+2)$. We simplify the first fraction and get $\frac{x y^{3}}{y-2}$. The expression $\frac{x^{2}}{y^{3}}+x^{-1}$ is transformed into $\frac{x^{3}+y^{3}}{x y^{3}}$. We factorize the sum of cubes. The expression $x y^{-2}-\frac{1}{y}+x^{-1}$ is transformed into $\frac{x^{2}-x y+y^{2}}{x y^{2}}$. The expression $\frac{(x-y)^{2}+4 x y}{1+\frac{y}{x}}$ is transformed into $\frac{x(x+y)^{2}}{x+y}=x(x+y)$. We factorize and simplify the last fraction, and get $y$.
We simplify the first term and get $\frac{y^{2}}{y-2}$. We subtract $y$ from this expression and get $\frac{2 y}{y-2}$.
Factoring $x y^{2018}+2 x y^{2017}=x y^{2017}(y+2)$ ..... 1 point
Factoring and factorizing $y^{2016}-4 y^{2014}=y^{2014}\left(y^{2}-4\right)=y^{2014}(y-2)(y+2)$ ..... 1 point
Transforming the first fraction into $\frac{x y^{3}}{y-2}$ ..... 1 point
Transforming $\left(\frac{x^{2}}{y^{3}}+x^{-1}\right) \text{into} \frac{x^{3}+y^{3}}{x y^{3}}$ ..... 1 point
Factorizing the sum of cubes ..... 1 point
Transforming $x y^{-2}-\frac{1}{y}+x^{-1} \text{into} \frac{x^{2}-x y+y^{2}}{x y^{2}}$ ..... 1 point
Transforming $\frac{(x-y)^{2}+4 x y}{1+\frac{y}{x}} \text{into} \frac{x(x+y)^{2}}{x+y}=x(x+y)$ ..... 1 point
Transforming $\frac{y^{2}+2 y}{y+2} \text{into} y$ ..... 1 point
Considering division, simplifying fractions to get $\frac{y^{2}}{y-2}$ from the first term ..... 1 point
Calculating $\frac{y^{2}}{y-2}-y=\frac{2 y}{y-2}$ ..... 1 point

## 18th Mathematics Knowledge Competition for Students of Secondary Technical and Vocational Schools

Preliminary Competition, March 15, 2018

## Solutions for the Second Year

A competitor who arrives at the solution by any correct method (even if the scoring guide does not provide for it) receives all possible points.

A correct method is considered to be any procedure that:

- sensibly takes into account the wording of the problem,
- leads to the solution of the problem,
- is mathematically correct and complete.

If an intermediate or final result can be recognized, guessed, read from a diagram, or calculated mentally, the competitor generally receives all the points provided for. If, however, the solution is guessed (cannot be arrived at without calculation), it is scored with 0 points.

A competitor who has only partially solved the problem, but from which the correct procedures for solving the problem are not visible, cannot receive more than half of the possible points.

The mark '*' next to the points means that the point or points can be awarded to the competitor for a correct procedure, even if the calculation is incorrect.

