3) 9x⁴ + 8x² - 1 = 0 4) x⁴ - 2x² + 1 = 0 5) (3x + 1)² - 4(3x + 1) - 5 = 0 6) 4(0,5x - 2)² + 7(0,5x - 2) - 2 = 0 7) (1-4x)² = 5(4x - 1) + 6 8) 3(x² - 4)² - 7 (x²-4)+4=0 9) 2(x² + x - 2) + 3(x² + x - 2) - 2 = 0 10) 5 – 10(1 − x²) + 5(x² - 1)² = 0 11) (x - 2)⁴ + 3(x - 2)² - 10 = 0 12) (x + 4)⁴ - 6(x + 4)² - 7 = 0

Решим уравнения:

1. $$9x^4 + 8x^2 - 1 = 0$$

   Пусть $$t = x^2$$, тогда $$t \ge 0$$.

   $$9t^2 + 8t - 1 = 0$$

   $$D = 8^2 - 4 \cdot 9 \cdot (-1) = 64 + 36 = 100$$

   $$t_1 = \frac{-8 + \sqrt{100}}{2 \cdot 9} = \frac{-8 + 10}{18} = \frac{2}{18} = \frac{1}{9}$$

   $$t_2 = \frac{-8 - \sqrt{100}}{2 \cdot 9} = \frac{-8 - 10}{18} = \frac{-18}{18} = -1$$ (не подходит, так как $$t \ge 0$$)

   $$x^2 = \frac{1}{9}$$

   $$x_1 = \frac{1}{3}, x_2 = -\frac{1}{3}$$

2. $$x^4 - 2x^2 + 1 = 0$$

   $$(x^2 - 1)^2 = 0$$

   $$x^2 - 1 = 0$$

   $$x^2 = 1$$

   $$x_1 = 1, x_2 = -1$$

3. $$(3x + 1)^2 - 4(3x + 1) - 5 = 0$$

   Пусть $$t = 3x + 1$$, тогда

   $$t^2 - 4t - 5 = 0$$

   $$D = (-4)^2 - 4 \cdot 1 \cdot (-5) = 16 + 20 = 36$$

   $$t_1 = \frac{4 + \sqrt{36}}{2 \cdot 1} = \frac{4 + 6}{2} = \frac{10}{2} = 5$$

   $$t_2 = \frac{4 - \sqrt{36}}{2 \cdot 1} = \frac{4 - 6}{2} = \frac{-2}{2} = -1$$

   $$3x + 1 = 5$$ или $$3x + 1 = -1$$

   $$3x = 4$$ или $$3x = -2$$

   $$x_1 = \frac{4}{3}, x_2 = -\frac{2}{3}$$

4. $$4(0.5x - 2)^2 + 7(0.5x - 2) - 2 = 0$$

   Пусть $$t = 0.5x - 2$$, тогда

   $$4t^2 + 7t - 2 = 0$$

   $$D = 7^2 - 4 \cdot 4 \cdot (-2) = 49 + 32 = 81$$

   $$t_1 = \frac{-7 + \sqrt{81}}{2 \cdot 4} = \frac{-7 + 9}{8} = \frac{2}{8} = \frac{1}{4}$$

   $$t_2 = \frac{-7 - \sqrt{81}}{2 \cdot 4} = \frac{-7 - 9}{8} = \frac{-16}{8} = -2$$

   $$0.5x - 2 = \frac{1}{4}$$ или $$0.5x - 2 = -2$$

   $$0.5x = \frac{9}{4}$$ или $$0.5x = 0$$

   $$x_1 = \frac{9}{2}, x_2 = 0$$

5. $$(1 - 4x)^2 = 5(4x - 1) + 6$$

   $$1 - 8x + 16x^2 = 20x - 5 + 6$$

   $$16x^2 - 28x = 0$$

   $$4x(4x - 7) = 0$$

   $$x_1 = 0, x_2 = \frac{7}{4}$$

6. $$3(x^2 - 4)^2 - 7(x^2 - 4) + 4 = 0$$

   Пусть $$t = x^2 - 4$$, тогда

   $$3t^2 - 7t + 4 = 0$$

   $$D = (-7)^2 - 4 \cdot 3 \cdot 4 = 49 - 48 = 1$$

   $$t_1 = \frac{7 + \sqrt{1}}{2 \cdot 3} = \frac{7 + 1}{6} = \frac{8}{6} = \frac{4}{3}$$

   $$t_2 = \frac{7 - \sqrt{1}}{2 \cdot 3} = \frac{7 - 1}{6} = \frac{6}{6} = 1$$

   $$x^2 - 4 = \frac{4}{3}$$ или $$x^2 - 4 = 1$$

   $$x^2 = \frac{16}{3}$$ или $$x^2 = 5$$

   $$x_1 = \frac{4}{\sqrt{3}}, x_2 = -\frac{4}{\sqrt{3}}, x_3 = \sqrt{5}, x_4 = -\sqrt{5}$$

7. $$2(x^2 + x - 2)^2 + 3(x^2 + x - 2) - 2 = 0$$

   Пусть $$t = x^2 + x - 2$$, тогда

   $$2t^2 + 3t - 2 = 0$$

   $$D = 3^2 - 4 \cdot 2 \cdot (-2) = 9 + 16 = 25$$

   $$t_1 = \frac{-3 + \sqrt{25}}{2 \cdot 2} = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2}$$

   $$t_2 = \frac{-3 - \sqrt{25}}{2 \cdot 2} = \frac{-3 - 5}{4} = \frac{-8}{4} = -2$$

   $$x^2 + x - 2 = \frac{1}{2}$$ или $$x^2 + x - 2 = -2$$

   $$x^2 + x - \frac{5}{2} = 0$$ или $$x^2 + x = 0$$

   Для $$x^2 + x - \frac{5}{2} = 0$$:

   $$D = 1^2 - 4 \cdot 1 \cdot (-\frac{5}{2}) = 1 + 10 = 11$$

   $$x_1 = \frac{-1 + \sqrt{11}}{2}, x_2 = \frac{-1 - \sqrt{11}}{2}$$

   Для $$x^2 + x = 0$$:

   $$x(x + 1) = 0$$

   $$x_3 = 0, x_4 = -1$$

8. $$5 - 10(1 - x^2) + 5(x^2 - 1)^2 = 0$$

   $$5 - 10 + 10x^2 + 5(x^4 - 2x^2 + 1) = 0$$

   $$10x^2 + 5x^4 - 10x^2 + 5 - 5 = 0$$

   $$5x^4 = 0$$

   $$x = 0$$

9. $$(x - 2)^4 + 3(x - 2)^2 - 10 = 0$$

   Пусть $$t = (x - 2)^2$$, тогда $$t \ge 0$$.

   $$t^2 + 3t - 10 = 0$$

   $$D = 3^2 - 4 \cdot 1 \cdot (-10) = 9 + 40 = 49$$

   $$t_1 = \frac{-3 + \sqrt{49}}{2 \cdot 1} = \frac{-3 + 7}{2} = \frac{4}{2} = 2$$

   $$t_2 = \frac{-3 - \sqrt{49}}{2 \cdot 1} = \frac{-3 - 7}{2} = \frac{-10}{2} = -5$$ (не подходит, так как $$t \ge 0$$)

   $$(x - 2)^2 = 2$$

   $$x - 2 = \sqrt{2}$$ или $$x - 2 = -\sqrt{2}$$

   $$x_1 = 2 + \sqrt{2}, x_2 = 2 - \sqrt{2}$$

10. $$(x + 4)^4 - 6(x + 4)^2 - 7 = 0$$

    Пусть $$t = (x + 4)^2$$, тогда $$t \ge 0$$.

    $$t^2 - 6t - 7 = 0$$

    $$D = (-6)^2 - 4 \cdot 1 \cdot (-7) = 36 + 28 = 64$$

    $$t_1 = \frac{6 + \sqrt{64}}{2 \cdot 1} = \frac{6 + 8}{2} = \frac{14}{2} = 7$$

    $$t_2 = \frac{6 - \sqrt{64}}{2 \cdot 1} = \frac{6 - 8}{2} = \frac{-2}{2} = -1$$ (не подходит, так как $$t \ge 0$$)

    $$(x + 4)^2 = 7$$

    $$x + 4 = \sqrt{7}$$ или $$x + 4 = -\sqrt{7}$$

    $$x_1 = \sqrt{7} - 4, x_2 = -\sqrt{7} - 4$$

Ответ: Решения уравнений выше.