a)  Với \(x \neq \&\text{nbsp}; \frac{1}{3}\), \(x \neq - \frac{1}{3}\). ta có:

\(P = \&\text{nbsp}; \left(\right. \frac{2 x}{3 x + 1} - 1 \left.\right) : \left(\right. 1 - \frac{8 x^{2}}{9 x^{2} - 1} \left.\right)\)

\(= \frac{2 x - 3 x - 1}{3 x + 1} : \frac{9 x^{2} - 1 - 8 x^{2}}{9 x^{2} - 1}\)

\(= \frac{- \left(\right. x + 1 \left.\right)}{3 x + 1} . \frac{9 x^{2} - 1}{x^{2} - 1}\)

\(= \frac{- \left(\right. x + 1 \left.\right)}{3 x + 1} . \frac{\left(\right. 3 x + 1 \left.\right) \left(\right. 3 x - 1 \left.\right)}{\left(\right. x + 1 \left.\right) \left(\right. x - 1 \left.\right)}\)

\(= \frac{1 - 3 x}{x - 1}\).
b) Thay \(x = 2\) vào biểu thức ta có:

     \(P = \frac{1 - 3.2}{2 - 1} = - 5\).

a) \(\frac{2 y - 1}{y} - \frac{2 x + 1}{x}\)

\(= \frac{x \left(\right. 2 y - 1 \left.\right)}{x y} - \frac{y \left(\right. 2 x + 1 \left.\right)}{x y}\)

\(= \frac{2 x y - x - 2 x y - y}{x y} = \frac{- x - y}{x y}\)
b) \(\frac{2 x}{3} : \frac{5}{6 x^{2}}\)

\(\frac{2 x}{3} : \frac{5}{6 x^{2}} = \frac{2 x}{3} . \frac{6 x^{2}}{5}\)

\(= \frac{4 x^{3}}{5}\)