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Question Number 51805 by azizullah last updated on 30/Dec/18

Commented by tanmay.chaudhury50@gmail.com last updated on 30/Dec/18

Commented by maxmathsup by imad last updated on 30/Dec/18
$the inequation is defined on R−{2} and (e) ⇔(((x+2)(x−2))/((x−2)^2 ))>0 ⇔ (x+2)(x−2)>0 and x≠2 ⇔ x ∈]−∞,−2[∪]2,+∞[.$
$t h e i n e q u a t i o n i s d e f i n e d o n R - {2} a n d (e) \Leftrightarrow \frac{(x + 2) (x - 2)}{{(x - 2)}^{2}} > 0 \Leftrightarrow$ $(x + 2) (x - 2) > 0 a n d x \neq 2 \Leftrightarrow x \in] - \infty, - 2 [\cup] 2, + \infty [.$
Commented by azizullah last updated on 31/Dec/18

$thanks so much dear sir .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 30/Dec/18

	$c r i t i c a l v a l u e o f x a r e - 2 a n d 2$ $1) x \neq 2 b e c a u s e a t x = 2 D_{r} = 0$ $2) w h e n x > 2 N_{r} > 0 a n d D_{r} > 0$ $\frac{N_{r}}{D_{r}} > 0 [g i v e n c o n d i t i o n \frac{x + 2}{x - 2} > 0 f u l f i l l]$ $3) 2 > x > - 2 [s a y x = 1 t h e n N_{r} > 0 D_{r} < 0 \frac{N_{r}}{D_{r}} < 0$ $s o g i v e n c o n d i t i o n \frac{x + 2}{x - 2} > 0 d o e s n o t f u l f i l l$ $3) x < - 2 [s a y x = - 3 N_{r} = - v e D_{r} = - v e b u t$ $\frac{N_{r}}{D_{r}} > 0 g i v e n c o n d i t i o n \frac{x + 2}{x - 2} > 0 f u l l f i l l$ $4) x = - 2 N_{r} = 0 D_{r} = - 4 \frac{N_{r}}{D_{r}} = 0$ $b u t g i v e n c o n d i t i o n d o e s n o t f u l f i l l$ $\frac{x + 2}{x - 2} > 0$ $s o s o l u t i o n s e t i s$ $(- \infty, - 2) \cup (2, + \infty)$
	Commented by azizullah last updated on 31/Dec/18

	$thanks dear sir .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 31/Dec/18

	$m o s t w e l c o m e . . .$
	Answered by afachri last updated on 30/Dec/18

	$‘ ‘ > 0^{″} means always possitive . It happen$ $when :$ $\frac{negative}{negative} or \frac{possitive}{possitive}$ $when (x - 2) is possitive, (x + 2) also possitive .$ $it happens when x :$ $x - 2 > 0$ $x > 2$ $when (x + 2) is negative, so is (x - 2) .$ $x + 2 < 0$ $x < - 2$ $set of solution {- 2 > x > 2, x \in R}$
	Commented by azizullah last updated on 31/Dec/18

	$thanks sir for solution of the given equation .$
	Commented by afachri last updated on 31/Dec/18

	${you}^{'} re welcome Azizullah$
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