Логарифмдік теңсіздіктерді шешу (Рустюмова 3.5.1А)

Шешуі:

ММЖ:

$${{x^2} - 7x \gt 0}$$ $${x(x - 7) \gt 0}$$ $${x = 0\quad x = 7}$$

$${3x + 11 \gt 0}$$ $${x \gt - \frac{{11}}{3}}$$

$${{x^2} - 7x \le 3x + 11}$$ $${{x^2} - 10x - 11 \le 0}$$ $${{x^2} - 10x - 11 = 0}$$ $${D = 100 + 44 = {{12}^2}}$$

$${{x_{1/2}} = \frac{{10 \pm 12}}{2}}$$ $${x = 11\quad x = - 1}$$

Жауабы: $x \in[-1,0) \cup[7 ; 11]$

Шешуі:

ММЖ:

$$\left\{\begin{array}{l}\frac{x+3}{x+4}\gt 0 \\ \frac{x+5}{x+2}\gt 0\end{array}\right.$$

$${\frac{{x + 3}}{{x + 4}} = 0}$$ $${x \ne - 4}$$ $${x = - 3}$$ $$(-\infty ;-4) \cup(-3 ;+\infty)$$

$${\frac{{x + 5}}{{x + 2}} = 0}$$ $${x \ne - 2\quad x = - 5}$$ $$(-\infty ;-5) \cup(-2 ;+\infty)$$

$$x \in(-\infty ;-5) \cup(-2 ;+\infty)$$

$${\frac{{x + 3}}{{x + 4}} \gt \frac{{x + 5}}{{x + 2}}}$$ $${\frac{{x + 3}}{{x + 4}} - \frac{{x + 5}}{{x + 2}} \gt 0}$$ $${\frac{{(x + 3)(x + 2) - (x + 5)(x + 4)}}{{(x + 4)(x + 2)}} \gt 0}$$ $${\frac{{{x^2} + 5x + 6 - \left( {{x^2} + 9x + 20} \right)}}{{(x + 4)(x + 2)}} \gt 0}$$

$${\frac{{{x^2} + 5x + 6 - {x^2} - 9x - 20}}{{(x + 4)(x + 2)}} \gt 0}$$ $${\frac{{ - 4x - 14}}{{(x + 4)(x + 2)}} \gt 0}$$ $${x + 4 \ne 0\quad x + 2 \ne 0}$$ $${x \ne - 4\quad x \ne - 2}$$ $${x = - \frac{7}{2}\quad }$$

Жауабы: $x \in(-\infty ;-5)$

Шешуі:

ММЖ: $$\left\{\begin{array}{l}2 x^2+4 x+10\gt 0 \\ x^2-4 x+3\gt 0\end{array} \Rightarrow\left\{\begin{array}{l}x^2+2 x+5\gt 0 \\ x^2-4 x+3\gt 0\end{array}\right.\right.$$

$$\begin{array}{l}x^2+2 x+5\gt 0 \\ x^2+2 x+5=0 \\ D=4-20=-16 \\ x \in(-\infty ;+\infty)\end{array}$$

$$\begin{array}{l}x^2-4 x+3\gt 0 \\ x^2-4 x+3=0 \\ D=16-12=4 \\ x_{1 / 2}=\frac{4 \pm 2}{2}=1; \; 3\end{array}$$ $$x \in(-\infty ; 1) \cup(3 ;+\infty)$$

$${2{x^2} + 4x + 10 \gt {x^2} - 4x + 3}$$ $${2{x^2} + 4x + 10 - {x^2} + 4x - 3 \gt 0}$$ $${{x^2} + 8x + 7 \gt 0}$$

$${{x^2} + 8x + 7 = 0}$$ $${D = 64 - 28 = 36}$$ $${{x_{1/2}} = \frac{{ - 8 \pm 6}}{2} = - 7;\, - 1}$$

Жауабы: $x \in(-\infty ;-7) \cup(-1 ; 1) \cup(3 ;+\infty)$

Шешуі:

ММЖ: $$\left\{ \begin{array}{l}\frac{4}{{x + 3}} \gt 0\\2 - x \gt 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x \in ( - 3; + \infty )\\x \in ( - \infty ;2)\end{array} \right.$$

$${x \in ( - 3;2)}$$

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

$${\frac{{{x^2} + x - 2}}{{x + 3}} \gt 0}$$ $${x \ne - 3;\quad {x^2} - 2 + x = 0}$$ $${D = 1 + 8 = 9}$$ $${{x_{1/2}} = \frac{{ - 1 \pm 3}}{2} = 1;\, - 2}$$

Жауабы: $x \in(-3 ;-2) \cup(1 ; 2)$

Шешуі:

ММЖ:

$$\frac{x-1}{x+2}\gt 0$$ $$x=1 \quad x \neq-2$$

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

$${{3^{{{\log }_2}\left( {\frac{{x - 1}}{{x + 2}}} \right)}} \lt {3^{ - 2}}}$$ $${{{\log }_2}\left( {\frac{{x - 1}}{{x + 2}}} \right) \lt - 2}$$ $${\frac{{x - 1}}{{x + 2}} \lt {2^{ - 2}}}$$

$${\frac{{x - 1}}{{x + 2}} - \frac{1}{4} \lt 0}$$ $${\frac{{4x - 4 - x - 2}}{{4(x + 2)}} \lt 0}$$ $${\frac{{3x - 6}}{{4(x + 2)}} \lt 0}$$

$${3x - 6 = 0;\quad x + 2 \ne 0}$$ $${x = 2;\quad x \ne - 2}$$ $$ x \in(-2 ; 2) $$

Жауабы: $x \in(1 ; 2)$

Шешуі:

ММЖ:

$${{x^2} - 5x + 6 \gt 0}$$ $${{x^2} - 5x + 6 = 0}$$ $${D = 25 - 24 = 1}$$

$${{x_{1/2}} = \frac{{5 \pm 1}}{2} = 3;\,2}$$ $$x \in(-\infty ; 2) \cup(3 ;+\infty)$$

$${{{\log }_{0,5}}\left( {{x^2} - 5x + 6} \right) \gt - 1}$$ $${{x^2} - 5x + 6 \lt {{0,5}^{ - 1}}}$$ $${{x^2} - 5x + 6 \lt 2}$$ $$x^2-5 x+4\lt 0 $$

$$\begin{array}{l} x^2-5 x+4=0 \\ D=25-16=9\end{array}$$ $${x_{1/2}} = \frac{{5 \pm 3}}{2} = 4;\,1$$ $$x \in(1 ; 4)$$

Жауабы: $x \in(1 ; 2) \cup(3 ; 4)$

Шешуі:

ММЖ:

$${{x^2} - 3x + 2 \gt 0}$$ $${{x^2} - 3x + 2 = 0}$$ $${D = 9 - 8 = 1}$$

$${{x_{1/2}} = \frac{{3 \pm 1}}{2} = 2;\,1}$$ $${x \in ( - \infty ;1) \cup (2; + \infty )}$$

$${{{\log }_{\sin \frac{\pi }{3}}}\left( {{x^2} - 3x + 2} \right) \ge 2}$$ $${{{\log }_{\frac{{\sqrt 3 }}{2}}}\left( {{x^2} - 3x + 2} \right) \ge 2}$$ $${{x^2} - 3x + 2 \le {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}$$ $${{x^2} - 3x + 2 - \frac{3}{4} \le 0}$$ $${{x^2} - 3x + 1\frac{1}{4} \le 0}$$

$${{x^2} - 3x + \frac{5}{4} = 0}$$ $${D = 9 - 4 \cdot \frac{5}{4} = 4}$$ $${{x_{1/2}} = \frac{{3 \pm 2}}{2} = \frac{5}{2};\,\frac{1}{2}}$$ $$x \in \left[ {\frac{1}{2};\frac{5}{2}} \right]$$

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

Шешуі:

ММЖ:

$${\frac{{2 - 3x}}{x} \gt 0}$$ $${x \ne 0,\quad x = \frac{2}{3}}$$

$$x \in\left(0 ; \frac{2}{3}\right)$$

$${{{\log }_{\frac{1}{3}}}\frac{{2 - 3x}}{x} \ge - 1}$$ $${\frac{{2 - 3x}}{x} \le {{\left( {\frac{1}{3}} \right)}^{ - 1}}}$$ $${\frac{{2 - 3x}}{x} - 3 \le 0}$$

$${\frac{{2 - 3x - 3x}}{x} \le 0}$$ $${\frac{{2 - 6x}}{x} \le 0}$$ $${x = \frac{1}{3},\quad x \ne 0}$$

$$x \in ( - \infty ;0) \cup \left[ {\frac{1}{3}; + \infty } \right)$$

Жауабы: $x \in \left[ {\frac{1}{3};\frac{2}{3}} \right)$

Шешуі:

ММЖ:

$$\begin{array}{lr}\frac{x-3}{x+3}\gt 0 \\ x-3=0 \quad x+3 \neq 0 \\ x=3 \qquad \;\; x \neq-3\end{array}$$

$$x \in(-\infty-3) \cup(3 ;+\infty)$$

$${{{\log }_{\frac{1}{4}}}\frac{{x - 3}}{{x + 3}} \ge - \frac{1}{2}}$$ $${\frac{{x - 3}}{{x + 3}} \le {{\left( {\frac{1}{4}} \right)}^{ - \frac{1}{2}}}}$$ $${\frac{{x - 3}}{{x + 3}} \le 2}$$

$${\frac{{x - 3}}{{x + 3}} - 2 \le 0}$$ $${\frac{{x - 3 - 2x - 6}}{{x + 3}} \le 0}$$ $${ - \frac{{x - 9}}{{x + 3}} \le 0}$$

$${ - x - 9 = 0;\quad x + 3 \ne 0}$$ $$x = - 9;\quad \quad \quad x \ne - 3$$

Жауабы: $x \in(-\infty ;-9] \cup(-3 ;+\infty)$

Шешуі:

ММЖ:$$\begin{array}{c}2 x+3\gt 0 \\ x\gt -\frac{3}{2}\end{array}$$

$${{{\log }_{\frac{1}{4}}}(2x + 3) \gt \frac{3}{2}}$$ $${2x + 3 \lt {{\left( {\frac{1}{4}} \right)}^{\frac{3}{2}}}}$$ $${2x + 3 \lt \frac{1}{8}}$$

$${2x \lt \frac{1}{8} - 3}$$ $${2x \lt - \frac{{23}}{8}}$$ $${x \lt - \frac{{23}}{{16}}}$$ $${x \lt - 1\frac{7}{{16}}}$$

Жауабы: $x \in\left(-\frac{3}{2} ;-1 \frac{7}{16}\right)$

Шешуі:

ММЖ:

$${{x^2} + 3x \gt 0}$$ $${x = 0;\quad x = - 3}$$

$$x \in(-\infty ;-3) \cup(0 ;+\infty)$$

$${2 - {{\log }_2}\left( {{x^2} + 3x} \right) \ge 0}$$ $${{{\log }_2}\left( {{x^2} + 3x} \right) \ge 2}$$ $${{x^2} + 3x \le 4}$$ $${x^2} + 3x - 4 \le 0$$

$${{x^2} + 3x - 4 = 0}$$ $${D = 9 + 16 = 25}$$ $${{x_{1/2}} = \frac{{ - 3 \pm 5}}{2} = 1;\, - 4}$$

Жауабы: $x \in[-4 ;-3) \cup(0 ; 1]$

Шешуі:

ММЖ:

$${{x^2} - 3x + 2 \gt 0}$$ $${{x^2} - 3x + 2 = 0}$$ $${x = 2;\quad x = 1}$$

$${x \in ( - \infty ;1) \cup (2; + \infty )}$$

$${{3^{{{\log }_2}\left( {{x^2} - 3x + 2} \right)}} \le 3}$$ $${{{\log }_2}\left( {{x^2} - 3x + 2} \right) \le 1}$$ $${{x^2} - 3x + 2 \le 2}$$ $${{x^2} - 3x \le 0}$$

$${x = 0;\quad x = 3}$$ $${x = 0;\quad x = 3}$$

Жауабы: $x \in[0 ; 1) \cup(2 ; 3]$

Шешуі:

$${4 - 9x \gt 0}$$ $${9x \lt 4}$$ $${x \lt \frac{4}{9}}$$

$${{{\log }_4}(4 - 9x) \lt 2}$$ $${4 - 9x \lt 16}$$ $${ - 9x \lt 12}$$ $${x \gt - \frac{4}{3}}$$

Жауабы: $x \in\left(-\frac{4}{3} ; \frac{4}{9}\right)$

Шешуі:

$${2{x^2} + x \gt 0}$$ $${x = 0;\quad x = - \frac{1}{2}}$$ $$x \in\left(-\infty ;-\frac{1}{2}\right) \cup(0 ;+\infty)$$

$${{{\log }_{\sqrt {10} }}\left( {2{x^2} + x} \right) \lt 2}$$ $${2{x^2} + x \lt 10}$$ $${2{x^2} + x - 10 \lt 0}$$ $${D = 1 + 4 \cdot 2 \cdot 10 = 81}$$ $${{x_{1/2}} = \frac{{ - 1 \pm 9}}{{2 \cdot 2}} = 2;\, - \frac{5}{2}}$$ $$x \in\left(-\frac{5}{2} ; 2\right)$$

Жауабы: $x \in\left(-\frac{5}{2} ;-\frac{1}{2}\right) \cup(0 ; 2)$

Шешуі:

$${{x^2} + 10x + 24 \gt 0}$$ $${x = - 6;\quad x = - 4}$$ $$x \in ( - \infty ; - 6) \cup ( - 4; + \infty )$$

$${6x + 36 \gt 0}$$ $${x \gt - 6}$$

$$x \in(-4 ;+\infty)$$

$${{{\log }_3}\left( {{x^2} + 10x + 24} \right) \le {{\log }_3}(6x + 36)}$$ $${{x^2} + 10x + 24 \le 6x + 36}$$ $${{x^2} + 4x - 12 \le 0}$$

$${x = - 6;\quad x = 2}$$ $$x \in [ - 6;2]$$

Жауабы: $x \in ( - 4;2]$

Шешуі:

ММЖ: $$x \neq-5 \quad x \neq \frac{1}{3}$$

$${{{\log }_{0,5}}{{(x + 5)}^2} \gt {{\log }_{\frac{1}{2}}}{{(3x - 1)}^2}}$$ $${{{(x + 5)}^2} \lt {{(3x - 1)}^2}}$$ $${{x^2} + 10x + 25 \lt 9{x^2} - 6x + 1}$$ $${ - 8{x^2} + 16x + 24 \lt 0}$$

$${{x^2} - 2x - 3 \gt 0}$$ $${x = 3;\quad x = - 1}$$ $$x \in ( - \infty ; - 1) \cup (3; + \infty )$$

Жауабы: $x \in ( - \infty ; - 5) \cup ( - 5; - 1) \cup (3; + \infty )$

Шешуі:

$${{x^2} - 3x + 4 \gt 0}$$ $${D = 9 - 16 = - 7}$$ $$x \in ( - \infty ; + \infty )$$

$${x + 1 \gt 0}$$ $${x \gt - 1}$$ $${x \in ( - 1;\, + \infty )}$$

$${\lg \sqrt {{x^2} - 3x + 4} \gt \lg \sqrt {x + 1} }$$ $${\sqrt {{x^2} - 3x + 4} \gt \sqrt {x + 1} }$$ $${{x^2} - 3x + 4 \gt x + 1}$$

$${{x^2} - 4x + 3 \gt 0}$$ $${x = 3;\quad x = 1}$$

Жауабы: $x \in ( - 1;1) \cup (3; + \infty )$

Шешуі:

$${2x - 1 \gt 0}$$ $${x \gt \frac{1}{2}}$$

$${\log _2}(2x - 1) \lt {\log _{{2^{ - \frac{1}{2}}}}}2$$ $${\log _2}(2x - 1) \lt {\log _2}{2^{ - 2}}$$

$${2x - 1 \lt \frac{1}{4}}$$ $${2x \lt \frac{5}{4}}$$ $${x \lt \frac{5}{8}}$$

Жауабы: ${x \in \left( {\frac{1}{2};\frac{5}{8}} \right)}$

Шешуі:

ММЖ: $$x\gt 0$$

$${{{\log }_3}x + {{\log }_{{3^{\frac{1}{2}}}}}x + {{\log }_{{3^{ - 1}}}}x \lt 6}$$ $${{{\log }_3}x + 2{{\log }_3}x - {{\log }_3}x \lt 6.}$$ $${{{\log }_3}x \lt 3}$$ $${x \lt 27}$$

Жауабы: $x \in(0 ; 27)$

Шешуі:

ММЖ: $$x \gt 0$$

$${\frac{1}{2}{{\log }_5}x - {{\log }_5}x \gt 2}$$ $${ - \frac{1}{2}{{\log }_5}x \gt 2}$$ $${{{\log }_5}x \lt - 4}$$ $$x\lt \frac{1}{625}$$

Жауабы: $x \in\left(0 ; \frac{1}{625}\right)$