142. Решите уравнение: 1) (2 + 3x)(4 - 6x + 9x²) - 3x(3x - 4)(3x + 4) = 10; 2) 16(1/2x - 2)(1/4x² + x + 4) - 2x(x - 6)² = 24x².

1) $$(2 + 3x)(4 - 6x + 9x^2) - 3x(3x - 4)(3x + 4) = 10$$ $$(2 + 3x)(4 - 6x + 9x^2) = 8 + 27x^3$$ $$3x(3x - 4)(3x + 4) = 3x(9x^2 - 16) = 27x^3 - 48x$$ $$8 + 27x^3 - (27x^3 - 48x) = 10$$ $$8 + 48x = 10$$ $$48x = 2$$ $$x = \frac{2}{48} = \frac{1}{24}$$ 2) $$16(\frac{1}{2}x - 2)(\frac{1}{4}x^2 + x + 4) - 2x(x - 6)^2 = 24x^2$$ $$16(\frac{1}{2}x - 2)(\frac{1}{4}x^2 + x + 4) = 16(\frac{1}{8}x^3 - 8) = 2x^3 - 128$$ $$2x(x - 6)^2 = 2x(x^2 - 12x + 36) = 2x^3 - 24x^2 + 72x$$ $$2x^3 - 128 - (2x^3 - 24x^2 + 72x) = 24x^2$$ $$2x^3 - 128 - 2x^3 + 24x^2 - 72x = 24x^2$$ $$-128 - 72x = 0$$ $$72x = -128$$ $$x = -\frac{128}{72} = -\frac{16}{9}$$