Решим уравнения:
$$x^2 - x = 0$$
$$x(x-1) = 0$$
$$x_1 = 0, x_2 = 1$$
$$y^2 - 4y = 0$$
$$y(y-4) = 0$$
$$y_1 = 0, y_2 = 4$$
$$5y^2 - y = 0$$
$$y(5y - 1) = 0$$
$$y_1 = 0, y_2 = \frac{1}{5}$$
$$10z^2 + 7z = 0$$
$$z(10z + 7) = 0$$
$$z_1 = 0, z_2 = -\frac{7}{10}$$
$$x(3x-5) - 4(3x-5) = 0$$
$$(3x-5)(x-4) = 0$$
$$x_1 = \frac{5}{3}, x_2 = 4$$
$$2y(9y-6) + 5(6-9y) = 0$$
$$18y^2 - 12y + 30 - 45y = 0$$
$$18y^2 - 57y + 30 = 0$$
$$6y^2 - 19y + 10 = 0$$
$$D = (-19)^2 - 4 \cdot 6 \cdot 10 = 361 - 240 = 121$$
$$y_1 = \frac{19 + 11}{12} = \frac{30}{12} = \frac{5}{2}$$, $$y_2 = \frac{19 - 11}{12} = \frac{8}{12} = \frac{2}{3}$$
Ответ: 1) $$x_1 = 0, x_2 = 1$$, 2) $$y_1 = 0, y_2 = 4$$, 3) $$y_1 = 0, y_2 = \frac{1}{5}$$, 4) $$z_1 = 0, z_2 = -\frac{7}{10}$$, 5) $$x_1 = \frac{5}{3}, x_2 = 4$$, 6) $$y_1 = \frac{5}{2}$$, $$y_2 = \frac{2}{3}$$