## Task B-2.6.

Solve the equation

$$
\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+4 x-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)=112-4 \sqrt{x}
$$
## First Solution.

First, note that $x>0, x \neq 1$ must hold.

1 point

By simplifying the expression in parentheses, we get

$$
\begin{array}{cc}
\sqrt{x}-\frac{1}{\sqrt{x}}=\frac{x-1}{\sqrt{x}}, & 1 \text { point } \\
\frac{\sqrt{x}+1}{\sqrt{x}-1}+4 x-\frac{\sqrt{x}-1}{\sqrt{x}+1}=4 x+\frac{(\sqrt{x}+1)^{2}-(\sqrt{x}-1)^{2}}{x-1}=4 x+\frac{4 \sqrt{x}}{x-1} . & 2 \text { points }
\end{array}
$$

Therefore, the given equation is equivalent to

$$
\frac{x-1}{\sqrt{x}} \cdot\left(4 x+\frac{4 \sqrt{x}}{x-1}\right)=112-4 \sqrt{x}
$$

It follows that

$$
\begin{array}{cc}
4(x-1) \sqrt{x}+4=112-4 \sqrt{x} & 2 \text { points } \\
4 x \sqrt{x}=108 & \\
x \sqrt{x}=27 . & 1 \text { point }
\end{array}
$$

Thus, $\sqrt{x}=3, x=9$.

Note: If the student wrote the condition for the solution at the beginning but did not emphasize when multiplying the equation by $\sqrt{x}$ and $x-1$ (or $\sqrt{x}-1$) that the next step yields an equivalent equation under the condition $x>0$ or $x \neq 1$, do not deduct a point.

Conversely, if the student emphasized when multiplying the equation by $\sqrt{x}$ and $x-1$ (or $\sqrt{x}-1$) that the next step yields an equivalent equation under the condition $x>0$ or $x \neq 1$, it is not necessary to write the condition at the beginning.

If neither is done, deduct 1 point.