Помогите пожалуйста решить

Ответ:

a) 6x⁴ + 15x² - 9 = 0

x² = t

6t² + 15t - 9 = 0

t = -3

t = 1/2

x² = -3

x² = 1/2

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

x = $\sqrt{2}$ /2

Ответ: x₁ = - $\sqrt{2}$ /2

x₂ = $\sqrt{2}$ /2

б) 2(x+3)⁴ - 5(x+3)² - 3 = 0

t = (x+3)²

2t² - 5t - 3 = 0

t = -1/2

t = 3

(x+3)² = -1/2

(x+3)² = 3

x = - $\sqrt{3}$ -3

x = $\sqrt{3}$ -3

Ответ: x₁ = - $\sqrt{3}$ -3; x₂ = $\sqrt{3}$ -3

в) 12/(x-1) - 8/(x+1) = 2

12/(x-1) - 8/(x+1) = 2, x≠1, x≠-1

12/(x-1) - 8/(x+1) - 2 = 0

$\frac{12(x+1) - 8(x-1) - 2(x-1)(x+1)}{(x-1)(x+1)} = 0$

$\frac{12x+12 - 8x + 8 - 2(x^2 - 1)}{(x-1)(x+1)} = 0$

$\frac{12x + 12 - 8x + 8 - 2 x^2 + 2}{(x-1)(x+1)} = 0$

$\frac{4x + 22 - 2 x^2}{(x-1)(x+1)} = 0$

4x + 22 - 2x² = 0

-2x² + 4x + 22 = 0

x² + 4x + 22 = 0

x = $\frac{-(-2) +- sqrt((-2)^2 - 4 * 1*(-11))}{2*1}$

x = (2± $\sqrt{48}$ )/2

x₁ = (2+ 4 $\sqrt{3}$ )/2

x₂ = (2- 4 $\sqrt{3}$ )/2

Ответ: x₁ = 1-2 $\sqrt{3}$ ; x₂ = 1 +2