9. Найдите значение выражения \((\frac{pq}{p+q}):(\frac{q}{p}-\frac{p}{q})\) при p = 3 – 2\(\sqrt{2}\), q = -2\(\sqrt{2}\)

Упростим выражение:

\((\frac{pq}{p+q}):(\frac{q}{p}-\frac{p}{q}) = \frac{pq}{p+q}:\frac{q^2-p^2}{pq} = \frac{pq}{p+q}\cdot\frac{pq}{q^2-p^2} = \frac{p^2q^2}{(p+q)(q-p)(q+p)} = \frac{p^2q^2}{(p+q)^2(q-p)}\)

Подставим значения p = 3 – 2\(\sqrt{2}\), q = -2\(\sqrt{2}\):

\(p^2 = (3-2\sqrt{2})^2 = 9 - 12\sqrt{2} + 8 = 17-12\sqrt{2}\)

\(q^2 = (-2\sqrt{2})^2 = 4\cdot2 = 8\)

\(p^2q^2 = (17-12\sqrt{2})\cdot8 = 136 - 96\sqrt{2}\)

\(p+q = 3-2\sqrt{2}-2\sqrt{2} = 3-4\sqrt{2}\)

\(q-p = -2\sqrt{2} - (3-2\sqrt{2}) = -2\sqrt{2}-3+2\sqrt{2} = -3\)

Выражение = \(\frac{136 - 96\sqrt{2}}{(3-4\sqrt{2})^2\cdot(-3)}\) = \(\frac{136 - 96\sqrt{2}}{(9-24\sqrt{2}+32)\cdot(-3)}\) = \(\frac{136 - 96\sqrt{2}}{(41-24\sqrt{2})\cdot(-3)}\) = \(\frac{136 - 96\sqrt{2}}{-123+72\sqrt{2}}\) - сложно упростить дальше без калькулятора.

Ответ: \(\frac{136 - 96\sqrt{2}}{-123+72\sqrt{2}}\)