B2. Simplify the expression $\left(\frac{x}{3}-\frac{x+1}{4}-\frac{1}{3 x}\right)^{-1}: \frac{12 x^{2}-24 x}{x^{2}-6 x+8}$ and calculate its value for $x=-2 \frac{3}{7}$.
B2. In the parentheses, we expand the fractions to the common denominator $12 x$. We get $\frac{4 x^{2}-3 x^{2}-3 x-4}{12 x}$ or $\frac{x^{2}-3 x-4}{12 x}$ and consider the negative exponent. The second fraction we factorize $\frac{12 x(x-2)}{(x-4)(x-2)}$ and simplify it, getting $\frac{12 x}{x-4}$. In the expression, we consider the division and get $\frac{12 x}{x^{2}-3 x-4} \cdot \frac{x-4}{12 x}$. We factorize the denominator of the first fraction $\frac{12 x}{(x-4)(x+1)} \cdot \frac{x-4}{12 x}$. We simplify the fractions and get $\frac{1}{x+1}$. We then calculate the value of the expression for $x=-2 \frac{3}{7}$. We get $\frac{1}{-2 \frac{3}{7}+1}$ and simplify $\frac{1}{-1 \frac{3}{7}}=\frac{1}{-\frac{10}{7}}$. We resolve the double fraction and get $-\frac{7}{10}$.

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Considering the negative exponent and division of fractions $\frac{12 x}{x^{2}-3 x-4} \cdot \frac{x-4}{12 x} \ldots \ldots \ldots \ldots . \ldots 1$ point

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