## Task A-1.1.

Let $x$ and $y$ be distinct real numbers such that $2xy + 1 \neq 0$ and let

$$
A=\frac{6x^2y^2+xy-1}{2xy+1} \quad \text{and} \quad B=\frac{x(x^2-1)-y(y^2-1)}{x-y}
$$

Determine which number is greater, $A$ or $B$.
## Solution.

By simplifying the given algebraic fractions, we get

$$
\begin{aligned}
A & =\frac{6 x^{2} y^{2}+3 x y-2 x y-1}{2 x y+1} \\
& =\frac{3 x y(2 x y+1)-(2 x y+1)}{2 x y+1} \\
& =\frac{(2 x y+1)(3 x y-1)}{2 x y+1} \\
& =3 x y-1
\end{aligned}
$$

and

$$
\begin{aligned}
B & =\frac{x\left(x^{2}-1\right)-y\left(y^{2}-1\right)}{x-y} \\
& =\frac{x^{3}-x-y^{3}+y}{x-y} \\
& =\frac{\left(x^{3}-y^{3}\right)-(x-y)}{x-y} \\
& =\frac{(x-y)\left(x^{2}+x y+y^{2}\right)-(x-y)}{x-y} \\
& =\frac{(x-y)\left(x^{2}+x y+y^{2}-1\right)}{x-y} \\
& =x^{2}+x y+y^{2}-1 .
\end{aligned}
$$

To determine which number is greater, we examine the difference between $A$ and $B$:

$$
\begin{aligned}
B-A & =\left(x^{2}+x y+y^{2}-1\right)-(3 x y-1) \\
& =x^{2}-2 x y+y^{2} \\
& =(x-y)^{2} \geqslant 0
\end{aligned}
$$

Since $x \neq y$, we conclude that $A<B$.