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

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

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

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

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

Пересечём множество решений: (4)

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

Ответ: x ∈ [-1;0]∪[4;5]