1+(1/2)log(3^1/2,(x+5)/(x+3))>=log(9,(x+1)^2)

1+(1/2)log(3^1/2,(x+5)/(x+3))>=log(9,(x+1)^2)
- Предмет:
  Алгебра
- Автор:
  zoemurphy
- 6 лет назад

Ответы 2

Не за что! :–)
- Автор:
  joyceali
- 6 лет назад
- 0
$1 + \frac{1}{2} \log_{ \sqrt{3} } { ( \frac{x+5}{x+3} ) } \geq \log_9 {(x+1)^2} \$ ;ОДЗ: $\left\{\begin{array}{l} \frac{x+5}{x+3} > 0 \ , \\\\ (x+1)^2 > 0 \ ; \end{array}ight$ $\left\{\begin{array}{l} x eq -3 \ , \\\\ \frac{x+5}{x+3} (x+3)^2 > 0 \ , \\\\ x eq -1 \ ; \end{array}ight$ $\left\{\begin{array}{l} x otin \{ -3 , -1 \} \ , \\\\ ( x + 3 ) ( x + 5 ) > 0 \ ; \end{array}ight$ $\left\{\begin{array}{l} x otin \{ -3 , -1 \} \ , \\ x otin [ -5 ; -3 ] \ ; \end{array}ight$ $x otin \{ [ -5 ; -3 ] \cup \{ -1 \} \} \ ;$ Решение: $1 + \log_{ \sqrt{3} } { \sqrt{ \frac{x+5}{x+3} } } \geq \log_{ \sqrt{9} } { \sqrt{ ( x + 1 )^2 } } \$ ; $\log_3 {3} + \log_3 { \frac{x+5}{x+3} } \geq \log_3 { |x+1| } \$ ; $\log_3 { ( 3 \cdot \frac{x+5}{x+3} ) } \geq \log_3 { |x+1| } \$ ; $3 \cdot \frac{x+5}{x+3} \geq |x+1| \$ ; $\left[\begin{array}{l} \left\{\begin{array}{l} x < -5 \ , \\ 3 ( x + 5 ) \leq -(x+1)(x+3) \ ; \end{array}ight \\\\ \left\{\begin{array}{l} -3 < x < -1 \ , \\ 3 ( x + 5 ) \geq -(x+1)(x+3) \ ; \end{array}ight \\\\ \left\{\begin{array}{l} x > -1 \ , \\ 3 (x+5) \geq (x+1)(x+3) \ ; \end{array}ight \end{array}ight$ $\left[\begin{array}{l} \left\{\begin{array}{l} x < -5 \ , \\ 3x + 15 + x^2 + 4x + 3 \leq 0 \ ; \end{array}ight \\\\ \left\{\begin{array}{l} -3 < x < -1 \ , \\ 3x + 15 + x^2 + 4x + 3 \geq 0 \ ; \end{array}ight \\\\ \left\{\begin{array}{l} x > -1 \ , \\ 3x + 15 \geq x^2 + 4x + 3 \ ; \end{array}ight \end{array}ight$ $\left[\begin{array}{l} \left\{\begin{array}{l} x < -5 \ , \\ x^2 + 7x + 18 \leq 0 \ ; \end{array}ight \\\\ \left\{\begin{array}{l} -3 < x < -1 \ , \\ x^2 + 7x + 18 \geq 0 \ ; \end{array}ight \\\\ \left\{\begin{array}{l} x > -1 \ , \\ x^2 + x - 12 \leq 0 \ ; \end{array}ight \end{array}ight$ $\left[\begin{array}{l} \left\{\begin{array}{l} x < -5 \ , \\ x^2 + 2 \cdot x \cdot \frac{7}{2} + ( \frac{7}{2} )^2 + 18 - \frac{49}{4} \leq 0 \ ; \end{array}ight \\\\ \left\{\begin{array}{l} -3 < x < -1 \ , \\ x^2 + 2 \cdot x \cdot \frac{7}{2} + ( \frac{7}{2} )^2 + 18 - \frac{49}{4} \geq 0 \ ; \end{array}ight \\\\ \left\{\begin{array}{l} x > -1 \ , \\ x^2 + 2 \cdot x \cdot \frac{1}{2} + ( \frac{1}{2} )^2 - 12 - \frac{1}{4} \leq 0 \ ; \end{array}ight \end{array}ight$ $\left[\begin{array}{l} \left\{\begin{array}{l} x < -5 \ , \\ ( x + \frac{7}{2} )^2 + 18 - 12 \frac{1}{4} \leq 0 \ ; \end{array}ight \\\\ \left\{\begin{array}{l} -3 < x < -1 \ , \\ ( x + \frac{7}{2} )^2 + 18 - 12 \frac{1}{4} \geq 0 \ ; \end{array}ight \\\\ \left\{\begin{array}{l} x > -1 \ , \\ ( x + \frac{1}{2} )^2 \leq \frac{49}{4} \ ; \end{array}ight \end{array}ight$ $\left[\begin{array}{l} \left\{\begin{array}{l} x < -5 \ , \\ ( x + \frac{7}{2} )^2 \leq - 5 \frac{3}{4} \ ; \end{array}ight \\\\ \left\{\begin{array}{l} -3 < x < -1 \ , \\ ( x + \frac{7}{2} )^2 \geq - 5 \frac{3}{4} \ ; \end{array}ight \\\\ \left\{\begin{array}{l} x > -1 \ , \\ | x + \frac{1}{2} | \leq \frac{7}{2} \ ; \end{array}ight \end{array}ight$ $\left[\begin{array}{l} \left\{\begin{array}{l} x < -5 \ , \\ x \in \emptyset \ ; \end{array}ight \\\\ \left\{\begin{array}{l} -3 < x < -1 \ , \\ x \in R \ ; \end{array}ight \\\\ \left\{\begin{array}{l} x > -1 \ , \\ ( x + 0.5 ) \in [ -3.5 ; 3.5 ] \ ; \end{array}ight \end{array}ight$ $\left[\begin{array}{l} -3 < x < -1 \ , \\\\ \left\{\begin{array}{l} x > -1 \ , \\ x \in [ -4 ; 3 ] \ ; \end{array}ight \end{array}ight$ $\left[\begin{array}{l} x \in ( -3 ; -1 ) \ , \\ x \in ( -1 ; 3 ] \ ; \end{array}ight$ $x \in ( -3 ; -1 ) \cup ( -1 ; 3 ] \equiv ( -3 ; 3 ]$ \ $\{ -1 \} \$ ;О т в е т : $x \in ( -3 ; -1 ) \cup ( -1 ; 3 ] \ .$
- Автор:
  mikaelaholloway
- 6 лет назад
- 0
Добавить свой ответ