Question

(x-2)⁴-13(x-2)²+36≤0

Answer 1

(x-2)^2=y

y^2-13y+36≤0

D=169-144=25

y1=(13+5)/2=9;(x-2)^2=9;x-2=+-3; x1=5;x2=-1

y2=(13-5)/2=4;(x-2)^2=4;x-2=+-2;x3=4;x3=0

+++[-1]----[0]++++[4]----[5]++++

Ответ x=[-1;0]U[4;5]

Answer 2

$(x-2)^{4} - 13(x-2)^{2} + 36 \leq 0 \\ \\ (x-2)^{2} = t \\ \\ t^{2} - 13t + 36 \leq 0 \\ D = 169 - 4 * 36 = 25 \\ \\ t_{1} = \dfrac{13 + 5}{2} = 9 \ ; \ t_{2} = \dfrac{13-5}{2} = 4 \\ \\ (t-9)(t-4)\leq 0 \ (1)\\ 4 \leq t \leq 9 \ \ \rightarrow \ 4 \leq (x-2)^{2} \leq 9 \ \ \rightarrow \ \ \left\{ \begin{aligned} (x-2)^{2} \ge 4\\ (x-2)^{2} \le 9 \\ \end{aligned} \right.$

$\left\{ \begin{gathered} x^{2} -4x + 4 \ge 4 \\ x^{2} - 4x + 4 \le 9 \\ \end{gathered} \right.$

$\left\{ \begin{gathered} x^{2} - 4x \ge 0 \ (a) \\ x^{2} - 4x - 5 \le 0 \ (b) \\ \end{gathered} \right.$

$(a): \ x(x - 4) \ge 0 \ (2) \\ x \in (-\infty ; 0]\cup[4;+\infty) \\ \\ (b): \ x^{2} - 4x - 5 \le 0 \\ D = 16 + 20 = 36 \\ \\ x_{1} = \dfrac{4 + 6}{2} = 5 \ ; \ x_{2} = \dfrac{4 - 6}{2} = -1 \\ \\ (x-5)(x+1)\le 0 \ (3) \\ x\in [-1;5]$

Пересечём множество решений: (4)

$x \in [-1;0]\cup [4;5]$

Ответ: x ∈ [-1;0]∪[4;5]