Бүтін рационал теңсіздіктер (Рустюмова-1.5.2Б)

Шешуі:

$${{{(x + 3)}^2}(x - 2){{(x + 5)}^3} = 0}$$ $${x = - 3,\quad x = 2,\quad x = - 5}$$

Жауабы: $x \in(-5 ;-3) \cup(-3 ; 2)$

Шешуі:

$${(8 - x){{(1 + x)}^2}{{(10 - x)}^3} = 0}$$ $${8 - x = 0,\quad 1 + x = 0,\quad 10 - x = 0}$$ $${x = 8,\quad \quad \,\,x = - 1,\quad \quad x = 10}$$

Жауабы: $x \in(-\infty ; 8] \cup[10 ;+\infty)$

Шешуі:

$${{{(x - 2)}^2}(x + 1)(x - 3) = 0}$$ $${x - 2 = 0,\quad x + 1 = 0,\quad x - 3 = 0}$$ $${x = 2\quad \quad \,\,\,x = - 1\quad \quad \,\,x = 3}$$

Жауабы: $x \in[-1 ; 3]$

Шешуі:

$${4{{\left( {x - \frac{1}{2}} \right)}^2}(x + 5)(x + 1) \le 0}$$ $${4{{\left( {x - \frac{1}{2}} \right)}^2}(x + 5)(x + 1) = 0}$$ $${x - \frac{1}{2} = 0,\quad x + 5 = 0,\quad x + 1 = 0,\quad }$$ $${x = \frac{1}{2}\quad \quad \quad x = - 5,\quad \quad x = - 1}$$

Жауабы: $x \in[-5 ;-1] \cup\left\{\frac{1}{2}\right\}$

Шешуі:

$${x{{(x - 1)}^2} = 0}$$ $${x = 0,\quad x = 1}$$

Жауабы: $x \in[0 ;+\infty)$

Шешуі:

$${ - {x^2} + 8x - 16 = 0}$$ $${ - {{(x - 4)}^2} = 0}$$ $${x = 4}$$

Жауабы: $\{4\}$

Шешуі:

$${{x^2}\left( {{x^2} + 8x + 12} \right) \ge 0}$$ $${{x^2}(x + 6)(x + 2) \ge 0}$$ $${{x^2}(x + 6)(x + 2) = 0}$$ $${x = 0,\quad x + 6 = 0,\quad x + 2 = 0}$$ $${\quad \quad \quad \,\,x = - 6,\quad \quad x = - 2}$$

Жауабы: $x \in(-\infty ;-6] \cup[-2 ;+\infty)$

Шешуі:

$${(x - 1)(x - 1)(x + 1) \le 0}$$ $${{{(x - 1)}^2}(x + 1) \le 0}$$ $${{{(x - 1)}^2}(x + 1) = 0}$$ $${x - 1 = 0,\quad x + 1 = 0}$$ $${x = 1,\quad \quad \quad x = - 1}$$

Жауабы: $x \in(-\infty ;-1] \cup\{1\}$

Шешуі:

$${(x + 3)^3}(x - 3)(x - 4){(x - 5)^2} = 0$$ $${x + 3 = 0,\quad x - 3 = 0,\quad x - 4 = 0,\quad x - 5 = 0}$$ $${x = - 3\quad \quad x = 3\quad \quad \,\,\,\,x = 4\quad \quad \,\,\,\,\,x = 5}$$

Жауабы: $$x \in ( - \infty ; - 3] \cup [3;4] \cup \{ 5\} $$

Шешуі:

$${(x - 3)(x + 3){{(x - 4)}^3}(x + 3) \le 0}$$ $${(x - 3){{(x + 3)}^2}{{(x - 4)}^3} \le 0}$$ $${(x - 3){{(x + 3)}^2}{{(x - 4)}^3} = 0}$$ $${x = 3,\quad x = - 3,\quad x = 4}$$

Жауабы: $x \in[3 ; 4] \cup\{-3\}$

Шешуі:

$${{{(3 - x)}^5}{{(x + 1)}^4}(x - 7) = 0}$$ $${x = 3,\quad x = - 1,\quad x = 7}$$

Жауабы: $x \in[3 ; 7] \cup\{-1\}$

Шешуі:

$${ - 9{x^2} + 6x + 6x - 4 \gt 0}$$ $${ - 3x(3x - 2) + 2(3x - 2) \gt 0}$$ $${ - (3x - 2)(3x - 2) \gt 0}$$

$${ - {{(3x - 2)}^2} \gt 0}$$ $${{{(3x - 2)}^2} \lt 0}$$

Жауабы: $\varnothing$

Шешуі:

$${\left( {7 - {x^2}} \right){{(x - 1)}^2}{{(x - 4)}^2} \ge 0}$$ $${\left( {7 - {x^2}} \right){{(x - 1)}^2}{{(x - 4)}^2} = 0}$$ $${x = \pm \sqrt 7 ,\quad x = 1,\quad x = 4}$$

Жауабы:

Шешуі:

$$(7 - x){(2 - x)^2}(x + 1) = 0$$ $${7 - x = 0,\quad 2 - x = 0,\quad x + 1 = 0}$$ $${x = 7\quad \quad \quad x = 2\quad \quad \,\,\,x = - 1}$$

Жауабы: $x \in[-1 ; 7]$

Шешуі:

$${ - \left( {{x^2} - 3x + 2} \right)\left( {{x^3} - 3{x^2}} \right)\left( {{x^2} - 4} \right) \le 0}$$ $${ - {x^2}\left( {{x^2} - 3x + 2} \right)(x - 3)(x - 2)(x + 2) \le 0}$$ $${ - {x^2}(x - 1)(x - 2)(x - 3)(x - 2)(x + 2) \le 0}$$ $${ - {x^2}(x - 1){{(x - 2)}^2}(x - 3)(x + 2) \le 0}$$ $${ - {x^2}(x - 1){{(x - 2)}^2}(x - 3)(x + 2) = 0}$$ $${x = 0,\quad x = 1,\quad x = 2,\quad x = 3,\quad x = - 2}$$

Жауабы: $x \in[-2 ; 1] \cup\{2\} \cup[3 ;+\infty)$

Шешуі:

$$(x - 2)(x - 4)(x - 2)(x + 2){(x - 2)^2} \ge 0$$ $${(x - 2)^4}(x + 2)(x - 4) \ge 0$$ $${(x - 2)^4}(x + 2)(x - 4) = 0$$ $$x = 2,\quad x = - 2,\quad x = 4$$

Жауабы: $x \in(-\infty ;-2] \cup[4 ;+\infty) \cup\{2\}$

Шешуі:

$${(x + 3){{(x(x - 1))}^2}(x - 2) \ge 0}$$ $${{x^2}{{(x - 1)}^2}(x - 2)(x + 3) = 0}$$ $${x = 0,\quad x = 1,\quad x = 2,\quad x = - 3}$$

Жауабы: $x \in(-\infty ;-3] \cup[2 ;+\infty) \cup\{0 ; 1\}$

Шешуі:

$${{{\left( {(x - 3)(x + 3)} \right)}^2}(x + 1)(x + 1)(x - 3)(x - 1) \le 0}$$ $${{{(x - 3)}^3}{{(x + 3)}^2}{{(x + 1)}^2}(x - 1) \le 0}$$ $${{{(x - 3)}^3}{{(x + 3)}^2}{{(x + 1)}^2}(x - 1) = 0}$$ $${x = 3,\quad x = - 3,\quad x = - 1,\quad x = 1}$$

Жауабы: $x \in[1 ; 3] \cup\{-3 ;-1\}$

Шешуі:

$${(x - 1)(3 - x){{(2 - x)}^2} = 0}$$ $${x = 1,\quad x = 3,\quad x = 2}$$

Жауабы: $x \in[1 ; 3]$

Шешуі:

$${\left( {{x^2} + 1} \right)(x + 1)(x - 5) \ge 0}$$ $${\left( {{x^2} + 1} \right)(x + 1)(x - 5) = 0}$$ $${x = - 1,\quad x = 5}$$

Жауабы: $x \in(-\infty ;-1] \cup[5 ;+\infty)$

Шешуі:

$${(x - 5)\left( {3{x^2} - x + 2} \right)(x - 5)(x + 5) \le 0}$$ $${{{(x - 5)}^2}(x + 5)\left( {3{x^2} - x + 2} \right) \le 0}$$ $${{{(x - 5)}^2}(x + 5)\left( {3{x^2} - x + 2} \right) = 0}$$ $${x = 5,\quad x = - 5}$$ $${3{x^2} - x + 2 = 0}$$ $${D = 1 - 24 = - 23 \lt 0}$$

Жауабы: $x \in(-\infty ;-5] \cup\{5\}$

Шешуі:

$${{x^4} + 7{x^2} - 9{x^2} - 63 \le 0}$$ $${{x^2}\left( {{x^2} + 7} \right) - 9\left( {{x^2} + 7} \right) \le 0}$$ $${\left( {{x^2} + 7} \right)\left( {{x^2} - 9} \right) \le 0}$$ $${\left( {{x^2} + 7} \right)(x - 3)(x + 3) = 0}$$ $${x = 3,\quad x = - 3,\quad {x^2} + 7 = 0}$$

Жауабы: $x \in[-3 ; 3]$

Шешуі:

$${\left( {{x^8} - 6{x^7} + 9{x^6}} \right) - \left( {{x^2} - 6x + 9} \right) \lt 0}$$ $${{x^6}\left( {{x^2} - 6x + 9} \right) - {{(x - 3)}^2} \lt 0}$$ $${{x^6}{{(x - 3)}^2} - {{(x - 3)}^2} \lt 0}$$ $${{{(x - 3)}^2}\left( {{x^6} - 1} \right) \lt 0}$$ $${{{(x - 3)}^2}\left( {{x^6} - 1} \right) = 0}$$ $${x = 3,\quad x = \pm 1}$$

Жауабы: $x \in(-1 ; 1)$

Шешуі:

$${\left( {3{x^2} - 5x + 8} \right)(x + 4) = 0}$$ $${3{x^2} - 5x + 8 = 0}$$ $${D = 25 - 96 = - 71 \lt 0}$$ $${x + 4 = 0}$$ $${x = - 4}$$

Жауабы: $x \in(-\infty ;-4)$

Шешуі:

$${{x^5} + {x^3} - {x^2} - 1 \lt 0}$$ $${\left( {{x^5} + {x^3}} \right) - \left( {{x^2} + 1} \right) \lt 0}$$ $${{x^3}\left( {{x^2} + 1} \right) - \left( {{x^2} + 1} \right) \lt 0}$$ $${\left( {{x^2} + 1} \right)\left( {{x^3} - 1} \right) \lt 0}$$

$${\left( {{x^2} + 1} \right)\left( {{x^3} - 1} \right) = 0}$$ $${\left( {{x^2} + 1} \right) \ne 0}$$ $${{x^3} - 1 = 0}$$ $${{x^3} = 1}$$

Жауабы: $x \in(-\infty ; 1)$

Шешуі:

$${\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right)\left( {{x^2} + 11} \right) \ge 0}$$ $${(x - 1)(x + 1)\left( {{x^2} + 1} \right)\left( {{x^2} + 11} \right) = 0}$$ $${x = 1,\quad x = - 1}$$

Жауабы: $x \in(-\infty ;-1] \cup[1 ;+\infty)$

Шешуі:

$${{x^2}(x - 3)\left( {{x^2} + 7} \right) \le 0}$$ $${{x^2}(x - 3)\left( {{x^2} + 7} \right) = 0}$$ $${x = 0,\quad x = 3}$$

Жауабы: $x \in(-\infty ; 3]$

Шешуі:

$${(x - 1)\left( {{x^2} - 1} \right)\left( {{x^3} - 1} \right)\left( {{x^4} - 1} \right) = 0}$$ $${x - 1 = 0,\quad {x^2} - 1 = 0,\quad {x^3} - 1 = 0,\quad {x^4} - 1 = 0}$$ $$x = 1,\quad \quad \quad x = \pm 1,\quad \quad {x^3} = 1,\quad \quad {x^4} = 1$$ $$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \,\,\,x = 1\quad \quad \,\,\,x = \pm 1$$

Жауабы: $\{-1 ; 1\}$

Шешуі:

$${{{(x - 1)}^5}(x + 2)\left( { - {x^2} + 2x - 10} \right) \lt 0}$$ $${ - {{(x - 1)}^5}(x + 2)\left( {{x^2} - 2x + 5} \right) \lt 0}$$ $${ - {{(x - 1)}^5}(x + 2)\left( {{x^2} - 2x + 5} \right) = 0}$$

$$x - 1 = 0,\quad x + 2 = 0$$ $$x = 1\quad \quad \,\,\,x = - 2$$ $${x^2} - 2x + 5 = 0$$ $${D = - 16 \lt 0}$$

Жауабы: $x \in(-\infty ;-2) \cup(1 ;+\infty)$

Шешуі:

$${(x - 3)\left( {3{x^2} - x + 2} \right)(x - 3)(x + 3) \ge 0}$$ $${{{(x - 3)}^2}(x + 3)\left( {3{x^2} - x + 2} \right) = 0}$$

$${x - 3 = 0,\quad x + 3 = 0}$$ $${x = 3,\quad \quad \,\,\,\,x = - 3}$$

Жауабы: $x \in[-3 ;+\infty)$

Шешуі:

$${x\left( {{x^2} - 4} \right)(x - 2)(x + 4)(x + 2)(x + 5)\left( {{x^2} + 1} \right) \le 0}$$ $${x{{(x - 2)}^2}{{(x + 2)}^2}(x + 4)(x + 5)\left( {{x^2} + 1} \right) \le 0}$$ $${x{{(x - 2)}^2}{{(x + 2)}^2}(x + 4)(x + 5)\left( {{x^2} + 1} \right) = 0}$$ $${x = 0,\quad x = 2,\quad x = - 2,\quad x = - 4,\quad x = - 5}$$

Жауабы: $x \in(-\infty ;-5] \cup[-4 ; 0] \cup\{2\}$

Шешуі:

$${(x - 3)\left( {{x^2} + 3} \right){{(x - 3)}^2} \le 0}$$ $${{{(x - 3)}^3}\left( {{x^2} + 3} \right) \le 0}$$ $${{{(x - 3)}^3}\left( {{x^2} + 3} \right) = 0}$$ $${x = 3,\quad {x^2} + 3 \gt 0}$$

Жауабы: $x \in(-\infty, 3]$

Шешуі:

$${ - \left( {{x^3} - 27} \right)\left( {{x^2} - 9} \right) \le 0}$$ $${ - (x - 3)\left( {{x^2} + 3x + 9} \right)(x - 3)(x + 3) \le 0}$$ $${ - {{(x - 3)}^2}(x + 3)\left( {{x^2} + 3x + 9} \right) = 0}$$ $${x - 3 = 0,\quad x + 3 = 0,\quad {x^2} + 3x + 9 = 0}$$ $${x = 3,\quad \quad \quad x = - 3,\quad \quad D \lt 0}$$

Жауабы: $x \in[-3 ;+\infty)$

Шешуі:

$${{{\left( {{x^2} - 1} \right)}^3}\left( {3{x^2} + 1} \right) + {{\left( {{x^2} - 1} \right)}^3}\left( {5{x^2} + 3x + 6} \right) \le 0}$$ $${{{\left( {{x^2} - 1} \right)}^3}\left( {3{x^2} + 1 + 5{x^2} + 3x + 6} \right) \le 0}$$ $${{{\left( {{x^2} - 1} \right)}^3}\left( {8{x^2} + 3x + 7} \right) \le 0}$$ $${{{\left( {{x^2} - 1} \right)}^3}\left( {8{x^2} + 3x + 7} \right) = 0}$$ $${\quad x = \pm 1\quad \quad \,\,\,8{x^2} + 3x + 7 \gt 0}$$ $${{x^2} = 1\quad \quad \quad D \lt 0}$$ $${x = \pm 1\quad \quad \,\,\,8{x^2} + 3x + 7 \gt 0}$$

Жауабы: $x \in[-1 ; 1]$