B2. The expression, $\mathrm{v}$ in which $x$ is a real number and $x \notin\{-3,-1,0,1,2,3,4\}$.

$$
\frac{x^{2}-4 x}{5 x-5} \cdot\left(\frac{x^{3}+1}{x^{2}+x}-1\right) \cdot\left(\left(1-\frac{3 x-3}{x^{2}+x-6}\right) \cdot\left(\frac{6}{x+3}-\frac{1}{x-2}\right)^{-1}-1\right)^{-1}
$$

Simplify it.
B2.

![](https://cdn.mathpix.com/cropped/2024_06_07_70800c8f4239f8164bacg-18.jpg?height=82&width=1742&top_left_y=2626&top_left_x=174)

Factorization of the sum of cubes .................................................................................................
Calculation $\frac{x^{3}+1}{x^{2}+x}-1=\frac{(x-1)^{2}}{x}$ ..... 1 point
Transformation $1-\frac{3 x-3}{x^{2}+x-6} \vee \frac{(x+1)(x-3)}{(x-2)(x+3)}$ ..... 1 point
Transformation $\left(\frac{6}{x-3}-\frac{1}{x-2}\right) \vee \frac{5(x-3)}{(x-2)(x+3)}$ ..... 1 point
Calculation $\left(1-\frac{3 x-3}{x^{2}+x-6}\right) \cdot\left(\frac{6}{x-3}-\frac{1}{x-2}\right)^{-1}=\frac{x+1}{5}$ ..... 1 point
Calculation $\left(\left(1-\frac{3 x-3}{x^{2}+x-6}\right) \cdot\left(\frac{6}{x-3}-\frac{1}{x-2}\right)^{-1}-1\right)^{-1}=\frac{5}{x-4}$ ..... 1 point
Result $x-1$ ..... 1 point