[tex] \frac{1}{x+2}- \frac{1}{x-2} \leq 1 [/tex] Решить при помощи метода интервалов.... Created by niksuperkrut. matematika-ru

нули функции: + - - +-----------(-2)------------[0]------------(2)------------ ////////////// //////////// ∈ ∞ ∪ {} ∪ ∞

tex] Решить при помощи метода интервалов...

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[tex] \frac{1}{x+2}- \frac{1}{x-2} \leq 1 [/tex] Решить при помощи метода интервалов...

Answers & Comments

Sophie155 $\frac{1}{x+2}- \frac{1}{x-2} \leq 1$
$\frac{x-2-(x+2)}{(x+2)(x-2)} \leq 1$
$\frac{x-2-x-2}{(x+2)(x-2)} \leq 1$
$\frac{-4}{(x+2)(x-2)} -1\leq 0$
$\frac{-4-(x^2-4)}{(x+2)(x-2)} \leq 0$
$\frac{-4-x^2+4}{(x+2)(x-2)} \leq 0$
$\frac{-x^2}{(x+2)(x-2)} \leq 0$
$\frac{x^2}{(x+2)(x-2)} \geq 0$
нули функции:
$x=0$
$x=2$
$x=-2$

+ - - +
-----------(-2)------------[0]------------(2)------------
////////////// ////////////

$x$ ∈ $(-$ ∞ $:-2)$ ∪ { ${0}$ } ∪ $(2;+$ ∞ $)$

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