6. Determine all solutions of the equation

(20)

$$
\left(x^{2}-5 x+5\right)^{2 \cdot 4^{x}-9 \cdot 2^{x}+4}=1
$$
Solution. We will consider three cases.

$1^{\circ}$ The base is equal to 1.

$$
x^{2}-5 x+5=1 \quad \Rightarrow \quad x_{1}=1, x_{2}=4
$$

Both solutions satisfy the original equation.

$2^{\circ}$ The exponent is equal to zero, and the base is different from zero.

$$
2 \cdot 4^{x}-9 \cdot 2^{x}+4=0
$$

With the substitution $t=2^{x}$, we get the quadratic equation

$$
2 t^{2}-9 t+4=0
$$

whose solutions are $t_{1}=4$ and $t_{2}=\frac{1}{2}$. From this, we get $x_{3}=2$ and $x_{4}=-1$. Since the base is different from zero for these solutions, both satisfy the original equation. (8 points) $3^{\circ}$ The exponent is an even number, and the base is equal to -1.

$$
x^{2}-5 x+5=-1 \quad \Rightarrow \quad x_{5}=2, x_{6}=3
$$

In both cases, the exponent is even, so these are solutions to the original equation.

All solutions to the equations are: $-1,1,2,3,4$.