22) \(\frac{1}{m}-...\)^2 = ... - ... + \(\frac{2}{35}\)mz^2+...

Вспоминаем формулу квадрата разности: (a-b)² = a² - 2ab + b².
Здесь a = \(\frac{1}{m}\). И 2ab = \(\frac{2}{35}\)mz^2, значит 2 * \(\frac{1}{m}\) * b = \(\frac{2}{35}\)mz^2, откуда b = \(\frac{1}{35}\)m^2z^2. Тогда a² = \(\frac{1}{m}\)^2 = \(\frac{1}{m^2}\). И b² = \(\frac{1}{35}m^2z^2\)^2 = \(\frac{1}{1225}\)m^4z^4
Тогда \(\frac{1}{m} - \frac{1}{35}m^2z^2\)^2 = \(\frac{1}{m^2}\) - \(\frac{2}{35}\)mz^2 + \(\frac{1}{1225}\)m^4z^4
Ответ: \(\frac{1}{m} - \frac{1}{35}m^2z^2\)^2 = \(\frac{1}{m^2}\) - \(\frac{2}{35}\)mz^2 + \(\frac{1}{1225}\)m^4z^4