7. Тип 7 № 6258 i Найдите значение выражения \(\frac{p^2-q^2}{(p-q)^2} - \frac{p^2+q^2}{(p+q)^2}\) при p = √б, и q = 2√2.

Выполним упрощение выражения:

$$\frac{p^2-q^2}{(p-q)^2} - \frac{p^2+q^2}{(p+q)^2} = \frac{(p-q)(p+q)}{(p-q)^2} - \frac{p^2+q^2}{(p+q)^2} = \frac{p+q}{p-q} - \frac{p^2+q^2}{(p+q)^2}$$ $$\frac{(p+q)^3 - (p^2+q^2)(p-q)}{(p-q)(p+q)^2} = \frac{p^3 + 3p^2q + 3pq^2 + q^3 - (p^3 - p^2q + q^2p - q^3)}{(p-q)(p+q)^2} =$$ $$\frac{p^3 + 3p^2q + 3pq^2 + q^3 - p^3 + p^2q - q^2p + q^3}{(p-q)(p+q)^2} = \frac{4p^2q + 2pq^2 + 2q^3}{(p-q)(p+q)^2} =$$ $$\frac{2q(2p^2 + pq + q^2)}{(p-q)(p+q)^2}$$

Подставим значения \(p = \sqrt{6}\) и \(q = 2\sqrt{2}\):

$$\frac{2 \cdot 2\sqrt{2}(2 \cdot 6 + \sqrt{6} \cdot 2\sqrt{2} + (2\sqrt{2})^2)}{(\sqrt{6} - 2\sqrt{2})(\sqrt{6} + 2\sqrt{2})^2} = \frac{4\sqrt{2}(12 + 4\sqrt{3} + 8)}{(\sqrt{6} - 2\sqrt{2})(\sqrt{6} + 2\sqrt{2})^2} =$$ $$\frac{4\sqrt{2}(20 + 4\sqrt{3})}{(\sqrt{6} - 2\sqrt{2})(\sqrt{6} + 2\sqrt{2})^2} = \frac{16\sqrt{2}(5 + \sqrt{3})}{(\sqrt{6} - 2\sqrt{2})(\sqrt{6} + 2\sqrt{2})^2}$$

Найдем значения в скобках:

$$(\sqrt{6} - 2\sqrt{2})(\sqrt{6} + 2\sqrt{2}) = 6 - 4 \cdot 2 = 6 - 8 = -2$$ $$\sqrt{6} + 2\sqrt{2} = \sqrt{2}(\sqrt{3} + 2)$$ $$\frac{16\sqrt{2}(5 + \sqrt{3})}{-2(\sqrt{6} + 2\sqrt{2})^2} = \frac{-8\sqrt{2}(5 + \sqrt{3})}{(\sqrt{6} + 2\sqrt{2})^2} = \frac{-8\sqrt{2}(5 + \sqrt{3})}{((\sqrt{3} + 2)\sqrt{2})^2} = \frac{-8\sqrt{2}(5 + \sqrt{3})}{2(\sqrt{3} + 2)^2} =$$ $$\frac{-4\sqrt{2}(5 + \sqrt{3})}{(3 + 4\sqrt{3} + 4)} = \frac{-4\sqrt{2}(5 + \sqrt{3})}{7 + 4\sqrt{3}}$$

Умножим числитель и знаменатель на сопряженное выражение знаменателя \(7 - 4\sqrt{3}\):

$$\frac{-4\sqrt{2}(5 + \sqrt{3})(7 - 4\sqrt{3})}{(7 + 4\sqrt{3})(7 - 4\sqrt{3})} = \frac{-4\sqrt{2}(35 - 20\sqrt{3} + 7\sqrt{3} - 12)}{49 - 16 \cdot 3} = \frac{-4\sqrt{2}(23 - 13\sqrt{3})}{49 - 48} =$$ $$\frac{-4\sqrt{2}(23 - 13\sqrt{3})}{1} = -4\sqrt{2}(23 - 13\sqrt{3}) = -92\sqrt{2} + 52\sqrt{6}$$

Ответ: \(-92\sqrt{2} + 52\sqrt{6}\)