Solve the following equations:

$$
\text { a) } \frac{\left\{\left(1+\frac{4}{x-2}\right)\left(x-4+4 x^{-1}\right)-\sqrt{3}:\left[1-(\sqrt{3}-1) 3^{-0.5}\right]\right\}(\sqrt{x}+2)^{-1}}{x^{-0.5}-2 x^{-1}}=7(x-1)-x^{2}
$$

b) The same as a), but on the right side, $7(x-1)-x^{2}$ is replaced by $x(10-x)-7$.

c) The same as a), but on the right side, $7(x-1)-x^{2}$ is replaced by $x(7-x)+1$.
Simplify the common left-hand side of all three equations as much as possible:

$$
\begin{gathered}
\frac{\left\{\frac{x+2}{x-2}\left(x-4+\frac{4}{x}\right)-\frac{\sqrt{3} \sqrt{3}}{\sqrt{3}-(\sqrt{3}-1)}\right\} \frac{1}{\sqrt{x}+2}}{\frac{1}{\sqrt{x}}-\frac{2}{x}}= \\
=\frac{\frac{x+2}{x-2}-\frac{x^{2}-4 x+4}{x} 3}{\frac{\sqrt{x}-2}{x} \cdot(\sqrt{x}+2)}=\frac{x(x+2)(x-2)^{2}-3 x^{2}(x-2)}{x(x-2)(x-4)}= \\
=\frac{x(x-2)\left[\left(x^{2}-4\right)-3 x\right]}{x(x-2)(x-4)}=\frac{x(x-2)(x-4)(x+1)}{x(x-2)(x-4)}
\end{gathered}
$$

Assuming $x \neq 0,2,4$, the left-hand side simplifies to: $x+1$.
a)

$$
x+1=7(x-1)-x^{2}
$$

which simplifies to

$$
x^{2}-6 x-8=0
$$

from which

$$
\left|x_{1}=4\right|, \quad\left|x_{2}=2\right|
$$

Since these two values of $x$ are excluded in (1), the equation under a) has no solution. b)

$$
x+1=x(10-x)-7
$$

which simplifies to

$$
x^{2}-9 x+8=0
$$

from which

$$
x_{1}=8, \quad x_{2}=1
$$

Both roots are solutions to the equation.
c)

$$
x+1=x(7-x)+1
$$

which simplifies to

$$
x^{2}-6 x=0
$$

from which

$$
\left|x_{1}=0\right|, \quad x_{2}=6
$$

Only $x_{2}=6$ is a solution to the equation, as $x_{1}=0$ is excluded in (1).