Question

Будь ласка допоможіть

Answer 1

Ответ:

1)   x1 = -1      x2 = 1

2)   x1 = -1       x2 = -1/3       x3 = 1/3       x4 = 1

3)    z1 = -1/2      z2 = 1

Объяснение:

1)

$x^{4}+8x^{2} -9=0\\\\t = x^{2} \\t^{2} +8t -9=0\\\\\left \{ {{t1+t2=-8} \atop {t1*t2 = -9}} \right. \\\\t1 = -9\\t2 = 1\\\\x^{2} = -9$↔  Нет решений в области действительных чисел.

$\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\x^{2} = 1\\x1 = -1 \\x2=1$

2)

$9x^{4} - 10x^{2} +1=0\\\\t = x^{2} \\9t^{2} - 10t +1 = 0\\\\D = 100-4*9 = 64\\t1 = \frac{10-8}{18} = \frac{1}{9} \\t2 = \frac{10+8}{18} = 1\\\\x^{2} = \frac{1}{9} \\x^{2} = 1\\\\x1 = -1\\x2 = -\frac{1}{3} \\x3 = \frac{1}{3} \\x4 = 1$

3)                                                     z ≠ 0, z ≠ -1

$\frac{z+2}{z} =\frac{5z+1}{z+1} \\(z+1)(z+2) = z(5z+1)\\z^{2} +3z+2 = 5z^{2} +z\\4z^{2} -2z-2 = 0\\2z^{2} -z-1 = 0\\\\D = 1+8= 9\\z1 = \frac{1-3}{4} = -\frac{1}{2} \\z2 = \frac{1+3}{4} =1$