Question

Решите уравнение.
x^4/(x+4)^2 =1

Answer 1

$\frac{ x^{4} }{(x+4) ^{2} }=1\\\\ \frac{ x^{4} }{(x+4) ^{2} } -1=0\\\\ \frac{ x^{4}-(x+4) ^{2} }{(x+4) ^{2} } =0\\\\ x^{4} -(x+4) ^{2} =0 , x \neq -4$
[x² - (x + 4)][x² + (x + 4)] = 0
x² - x - 4 = 0                                                            x² + x + 4 = 0
D = (- 1)² - 4 * 1 * (- 4) = 1 + 16 = 17            D = 1² - 4 * 1 * 4 = 1 - 16 = - 15 < 0
$x_{1} = \frac{1+ \sqrt{x}17}{2}$                                      D < 0 -  решений нет

$x_{2} = \frac{1- \sqrt{17} }{2}$