Найдите значение алгебраической дроби, предварительно сократив ее:
а) $\frac{40x^2 - 5xy}{y^2 - 8xy}$ при x = 2, y = 10;
б) $\frac{21a^2 - 12ab}{20b^2 - 35ab}$ при a = 10, b = −3;
в) $\frac{15c^2 - 10cd}{8d^2 - 12cd}$ при c = −6, d = 4;
г) $\frac{25z^2 - 20zt}{16t^2 - 20zt}$ при z = −3, t = −2.
$\frac{40x^2 - 5xy}{y^2 - 8xy} = \frac{5x(8x - y)}{y(y - 8x)} = -\frac{5x(y - 8x)}{y(y - 8x)} = -\frac{5x}{y}$
при x = 2, y = 10:
$-\frac{5x}{y} = -\frac{5 * 2}{10} = -\frac{10}{10} = -1$
$\frac{21a^2 - 12ab}{20b^2 - 35ab} = \frac{3a(7a - 4b)}{5b(4b - 7a)} = -\frac{3a(7a - 4b)}{5b(7a - 4b)} = -\frac{3a}{5b}$
при a = 10, b = −3:
$-\frac{3 * 10}{5 * (-3)} = -\frac{30}{-15} = 2$
$\frac{15c^2 - 10cd}{8d^2 - 12cd} = \frac{5c(3c - 2d)}{4d(2d - 3c)} = -\frac{5c(2d - 3c)}{4d(2d - 3c)} = -\frac{5c}{4d}$
при c = −6, d = 4:
$-\frac{5 * (-6)}{4 * 4} = \frac{5 * 3}{2 * 4} = \frac{15}{8} = 1\frac{7}{8}$
$\frac{25z^2 - 20zt}{16t^2 - 20zt} = \frac{5z(5z - 4t)}{4t(4t - 5z)} = -\frac{5z(4t - 5z)}{4t(4t - 5z)} = -\frac{5z}{4t}$
при z = −3, t = −2:
$-\frac{5 * (-3)}{4 * (-2)} = -\frac{-15}{-8} = -1\frac{7}{8}$
Пожауйста, оцените решение
