## Task A-1.1.

Determine the real number $a$ so that $x=\frac{1}{2}$ is a solution to the equation

$$
\left(\frac{x+1}{1-x}-\frac{1-x}{x+1}-\frac{4 x^{2}}{x^{2}-1}\right)\left(\frac{1}{a^{3}+a^{2}}-\frac{1-a}{a^{2}}-1\right)=\frac{1}{2}
$$
## First Solution.

First, we simplify the expressions in parentheses:

$$
\begin{array}{rlrl}
\frac{x+1}{1-x}-\frac{1-x}{x+1}-\frac{4 x^{2}}{x^{2}-1}= & \frac{(x+1)^{2}-(1-x)^{2}+4 x^{2}}{1-x^{2}}=\frac{4 x^{2}+4 x}{1-x^{2}}=\frac{4 x}{1-x}, & & 1 \text{ point} \\
\frac{1}{a^{3}+a^{2}}-\frac{1-a}{a^{2}}-1 & =\frac{1}{a^{2}(a+1)}-\frac{1-a}{a^{2}}-1 & \\
& =\frac{1-(1-a)(a+1)-a^{2}(a+1)}{a^{2}(a+1)} \\
& =\frac{a^{2}-a^{2}(a+1)}{a^{2}(a+1)}=\frac{-a^{3}}{a^{2}(a+1)}=-\frac{a}{a+1} . & 1 \text{ point}
\end{array}
$$

Thus, the given equation becomes

$$
\frac{4 x}{1-x} \cdot\left(-\frac{a}{a+1}\right)=\frac{1}{2}
$$

Since the value of the expression $\frac{4 x}{1-x}$ for $x=\frac{1}{2}$ is $\frac{4 \cdot \frac{1}{2}}{1-\frac{1}{2}}=4$,

for $x=\frac{1}{2}$ to be a solution of this equation, it must be that $4\left(-\frac{a}{a+1}\right)=\frac{1}{2}$.

1 point

Solving this equation for $a$ yields

$$
-\frac{a}{a+1}=\frac{1}{8}, \quad-8 a=a+1, \quad-9 a=1
$$

so $a=-\frac{1}{9}$.

1 point