Предмет: Алгебра, автор: toraliali65

x(x+3)(x-4)/(x+1)(x+2)≤0

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Автор ответа: Universalka

$\displaystyle\bf\\\frac{x(x+3)(x-4)}{(x+1)(x+2)} \leq 0$

Нули функции , стоящей в левой части неравенства :

$\displaystyle\bf\\x_{1} =0\\\\x+3=0 \ \ \ \Rightarrow \ \ \ x_{2} =-3\\\\x-4=0 \ \ \ \Rightarrow \ \ \ x_{3} =4$

Числа , не входящие в область определения функции , так как они обращают знаменатель в ноль :

$\displaystyle\bf\\x+1\neq 0 \ \ \ \Rightarrow \ \ \ x\neq -1\\\\x+2\neq 0 \ \ \ \Rightarrow \ \ \ x\neq -2$

Отметим нули функции и числа, не входящие в область определения на числовой прямой и расставим знаки в каждом из промежутков .

Неравенство не строгое, значит нули функции будут закрашенными точками , а две другие точки незакрашенные .

- - - - - [-3]+ + + + + (-2)- - - - - (-1)+ + + + + [0]- - - - - [4]+ + + + +

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Ответ : x ∈ ( - ∞ ; - 3] ∪ ( - 2 ; - 1) ∪ [0 ; 4]

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x(x+3)(x-4)/(x+1)(x+2)≤0​

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x(x+3)(x-4)/(x+1)(x+2)≤0