If $x = -1$ and $x = -2$ are the extreme points of $f(x) = \alpha\log|x| + \beta x^2 + x$ then

At extreme points, the derivative equals zero.

Given: $f(x) = \alpha\log|x| + \beta x^2 + x$

The derivative of $\log|x|$ is $\frac{1}{x}$

$f'(x) = \frac{\alpha}{x} + 2\beta x + 1$

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Since $x = -1$ is an extreme point, $f'(-1) = 0$:

$\frac{\alpha}{-1} + 2\beta(-1) + 1 = 0$

$-\alpha - 2\beta + 1 = 0$

$\alpha + 2\beta = 1$ ...(Equation 1)

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Since $x = -2$ is an extreme point, $f'(-2) = 0$:

$\frac{\alpha}{-2} + 2\beta(-2) + 1 = 0$

$-\frac{\alpha}{2} - 4\beta + 1 = 0$

Multiplying by 2:

$-\alpha - 8\beta + 2 = 0$

$\alpha + 8\beta = 2$ ...(Equation 2)

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From Equation 1: $\alpha + 2\beta = 1$

From Equation 2: $\alpha + 8\beta = 2$

Subtracting Equation 1 from Equation 2:

$6\beta = 1$

$\beta = \frac{1}{6}$

Substituting into Equation 1:

$\alpha + 2\left(\frac{1}{6}\right) = 1$

$\alpha + \frac{1}{3} = 1$

$\alpha = \frac{2}{3}$

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Therefore, $\alpha = \frac{2}{3}$ and $\beta = \frac{1}{6}$