Task B-1.6. Determine the value of the real parameter $a$ so that the solution of the equation

$$
\frac{2 a+x}{2-x}-\frac{2 a-x}{2+x}=\frac{4 a}{4-x^{2}}
$$

is less than or equal to 1.
## Solution.

For the given equation to make sense, it must be $2-x \neq 0$ and $2+x \neq 0$, i.e., $x \neq \pm 2$.

(1 point)

Then we have

$$
\begin{aligned}
\frac{2 a+x}{2-x}-\frac{2 a-x}{2+x} & =\frac{4 a}{4-x^{2}} \\
\frac{2 a+x}{2-x}-\frac{2 a-x}{2+x} & =\frac{4 a}{(2-x)(2+x)} / \cdot(2-x)(2+x) \neq 0 \\
(2 a+x)(2+x)-(2 a-x)(2-x) & =4 a \\
4 a+2 a x+2 x+x^{2}-\left(4 a-2 a x-2 x+x^{2}\right) & =4 a \\
4 a x+4 x & =4 a
\end{aligned}
$$

Factoring the left side of the equation and dividing, under the condition $a+1 \neq 0 (a \neq -1)$,

$$
\begin{aligned}
4 x(a+1) & =4 a /:(a+1) \neq 0 \\
x & =\frac{a}{a+1}, \quad a \neq -1
\end{aligned}
$$

Since $x$ must not be 2 or -2, we have $\frac{a}{a+1} \neq 2$, which implies $a \neq -2$.

Furthermore, $\frac{a}{a+1} \neq -2$, which implies $a \neq -\frac{2}{3}$.

Finally, the solution to the given equation is $x=\frac{a}{a+1}$ for $a \neq -2, -1, -\frac{2}{3}$.

We need to check when this is less than or equal to 1. We solve the inequality

$$
\begin{aligned}
\frac{a}{a+1} & \leq 1 \\
\frac{a}{a+1}-1 & \leq 0 \\
\frac{-1}{a+1} & \leq 0
\end{aligned}
$$

The last inequality is equivalent to the inequality $a+1>0$, whose solution is $a>-1$.

(1 point)

Finally, the solution to the equation will be less than or equal to 1 for $a \in (-1, \infty) \setminus \left\{-\frac{2}{3}\right\}$. (1 point)