282. Solve the equation:

$$
x^{4}-5 x^{3}+8 x^{2}-5 x+1=0
$$
$\triangle$ Let's divide this equation term by term by $x^{2}$:

$$
x^{2}-5 x+8-\frac{5}{x}+\frac{1}{x^{2}}=0, \quad\left(x^{2}+\frac{1}{x^{2}}\right)-5\left(x+\frac{1}{x}\right)+8=0
$$

Let $x+\frac{1}{x}=y$. Square this equality:

$$
x^{2}+2+\frac{1}{x^{2}}=y^{2}, \quad x^{2}+\frac{1}{x^{2}}=y^{2}-2
$$

The equation takes the following form:

$$
y^{2}-2-5 y+8=0, \quad y^{2}-5 y+6=0
$$

From here, $y_{1}=2, y_{2}=3$. Now it is not difficult to find $x$.

Answer: $1,1,(3 \pm \sqrt{5}) / 2$.