Question

(x + 2)⁴ - (x²+4x) - 6 = 0​

Answer 1

$></p> <p>Пусть x²+4x = t, тогда (x+2)² = x²+4x+4 = t+4</p> <p><img src=$

Теперь сделаем обратную замену :

Для первого t

$x^{2} +4x=-2\\x^{2} +4x+2=0\\D=16-8=8=(2\sqrt{2} )^{2} \\x1=\frac{-4+2\sqrt{2} }{2} =\frac{2(-2+\sqrt{2}) }{2} =-2+\sqrt{2}\\x2=\frac{-4-2\sqrt{2} }{2} =\frac{2(-2-\sqrt{2}) }{2} =-2-\sqrt{2}$

Для второго t

$x^{2} +4x=-5\\x^{2} +4x+5=0\\D=16-20=-4\\D$

∅

Ответ : -2+√2 , -2-√2

Answer 2

To solve the equation (x + 2)⁴ - (x² + 4x) - 6 = 0, we can follow these steps:

Step 1: Expand the equation. (x + 2)⁴ - (x² + 4x) - 6 = 0 (x + 2)(x + 2)(x + 2)(x + 2) - (x² + 4x) - 6 = 0 (x + 2)(x + 2)(x + 2)(x + 2) - x² - 4x - 6 = 0

Step 2: Simplify the equation. Let's simplify the left side of the equation: (x + 2)(x + 2) = (x² + 4x + 4) So we have: (x + 2)(x + 2)(x + 2)(x + 2) - x² - 4x - 6 = 0 (x² + 4x + 4)(x² + 4x + 4) - x² - 4x - 6 = 0 (x² + 4x + 4)² - x² - 4x - 6 = 0 (x⁴ + 8x³ + 24x² + 32x + 16) - x² - 4x - 6 = 0 x⁴ + 8x³ + 24x² + 32x + 16 - x² - 4x - 6 = 0 x⁴ + 8x³ + 23x² + 28x + 10 = 0

Now we have a quartic equation, which can be difficult to solve directly. However, there are numerical methods and approximation techniques that can be used to find approximate solutions.