Вычислите $\frac{2x^2 - 2y^2}{(x + y)(x - y)}$, если:
а) x = 2, y = 3;
б) $x = \frac{3}{2}, y = \frac{1}{3}$;
в) x = −2, y = 0;
г) x = 1,3, y = −0,5.
$\frac{2x^2 - 2y^2}{(x + y)(x - y)} = \frac{2 * 2^2 - 2 * 3^2}{(2 + 3)(2 - 3)} = \frac{2 * 4 - 2 * 9}{5 * (-1)} = \frac{8 - 18}{-5} = \frac{-10}{-5} = 2$
$\frac{2x^2 - 2y^2}{(x + y)(x - y)} = \frac{2 * (\frac{3}{2})^2 - 2 * (\frac{1}{3})^2}{(\frac{3}{2} + \frac{1}{3})(\frac{3}{2} - \frac{1}{3})} = \frac{2 * \frac{9}{4} - 2 * \frac{1}{9}}{(\frac{9}{6} + \frac{2}{6})(\frac{9}{6} - \frac{2}{6})} = \frac{\frac{9}{2} - \frac{2}{9}}{\frac{11}{6} * \frac{7}{6}} = \frac{\frac{81}{18} - \frac{4}{18}}{\frac{77}{36}} = \frac{77}{18} * \frac{36}{77} = \frac{1}{1} * \frac{2}{1} = 2$
$\frac{2x^2 - 2y^2}{(x + y)(x - y)} = \frac{2 * (-2)^2 - 2 * 0^2}{(-2 + 0)(-2 - 0)} = \frac{2 * 4}{-2 * (-2)} = \frac{8}{4} = 2$
$\frac{2x^2 - 2y^2}{(x + y)(x - y)} = \frac{2 * 1,3^2 - 2 * (-0,5)^2}{(1,3 - 0,5)(1,3 + 0,5)} = \frac{2 * 1,69 - 2 * 0,25}{0,8 * 1,8} = \frac{2 * (1,69 - 0,25)}{0,8 * 1,8} = \frac{2 * 1,44}{1,44} = 2$
Пожауйста, оцените решение
