Question

(x²-x+-1)(x²-x-7)≤-5 розвязати нерівність

Answer 1

$(x^2-x-1)(x^2-x-7)\leq -5\\ \\ t=x^2-x\\ \\ (t-1)(t-7)\leq-5\\ \\ t^2-8t+7 \leq -5\\ \\ t^2-8t+12\leq 0\\ \\ D=64-48=16\\ \\ t_{1} =(8+4)/2=6\\ \\ t_{2} =(8-4)/2=2\\ \\ (t-6)(t-2)\leq 0\\ \\ (x^2-x-6)(x^2-x-2)\leq 0\\ \\ D=25;x_{1} =(1+5)/2=3;x_{2} =(1-5)/2=-2\\ \\ D=9;x_{1} =(1+3)/2=2;x_{2} =(1-3)/2=-1\\ \\ +++[-2]---[-1]+++[2]---[3]+++\\ \\ x\in[-2;-1]U[2;3]$

Answer 2

$(x^{2} - x -1)(x^{2} - x- 7) \le -5 \\ \\ x^{2} - x = t \\ \\ (t - 1)(t-7) \le -5 \\ t^{2} -8t+ 12 \le 0 \\ D = 64 - 4 * 12 = 16 \\ \\ t_{1} = \dfrac{8 + 4}{2} = 6 \ ; \ t_{2} = \dfrac{8 - 4}{2} = 2 \\ \\ (t-6)(t-2) \le 0 \ (1) \\ \\ 2 \le t \le 6 \ \ \rightarrow \ \ 2 \le x^{2} - x \le 6 \ \ \rightarrow \ \ \begin{equation*} \begin{cases} x^{2} - x \ge 2 \\ x^{2} - x \le 6 \\ \end{cases}\end{equation*}$

$\begin{equation*} \begin{cases} x^{2} - x -2 \ge 0 \ (a) \\ x^{2} - x -6 \le 0 \ (b) \\ \end{cases}\end{equation*}$

$(a): \ x^{2} - x - 2 \ge 0 \\ D = 1 + 8 = 9 \\ \\ x_{1} = \dfrac{1 + 3}{2} = 2 \ ; \ x_{2} = \dfrac{1 - 3}{2} = - 1 \\ \\ (x-2)(x+1) \ge 0 \ (2) \\ x\in (-\infty ; -1]\cup [2;+\infty) \\ \\(b): \ x^{2} - x - 6 \le 0 \\ D = 1 + 24 = 25 \\ \\ x_{1} = \dfrac{1 + 5}{2} = 3 \ ; \ t_{2} = \dfrac{1 -5}{2} = - 2 \\ \\ (x-3)(x+2) \le 0 \ (3) \\ x\in [-2;3]$

Пересечём множество решений (4):

$x\in[-2;-1]\cup [2;3]$

Ответ: x∈[-2;-1]∪[2;3]