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Question Number 33767 by artibunja last updated on 23/Apr/18

	Answered by MJS last updated on 23/Apr/18
	$(x−3)^2 >0 with x∈R\{3} ⇒ multiplication with denominator doesn′t change sign x^2 −7∣x∣+10<0 case 1.: x<0 x^2 +7x+10<0 (x+2)(x+5)<0 ⇒ [(x+2)<0∧(x+5)>0]∨[(x+2)>0∧(x+5)<0] case 1.1.: x<−2∧x>−5 ⇒ x∈]−5;−2[ case 1.2.: x>−2∧x<−5 ⇒ no solution case 2.: x≥0 x^2 −7x+10<0 (x−2)(x−5)<0 ⇒ [(x−2)<0∧(x−5)>0]∨[(x−2)>0∧(x−5)<0] case 2.1.: x<2∧x>5 ⇒ no solution case 2.2.: x>2∧x<5 ⇒ x∈]2;5[ at x=3 the limit of the function is −∞<0 solution is x∈]−5;−2[ ∩ ]2;5[ or ∣x∣∈]2;5[ or 2<∣x∣<5$
	${(x - 3)}^{2} > 0 with x \in R ∖ {3} \Rightarrow multiplication$ $with denominator {doesn}^{'} t change sign$ $x^{2} - 7 ∣ x ∣ + 10 < 0$ $case 1 . :$ $x < 0$ $x^{2} + 7 x + 10 < 0$ $(x + 2) (x + 5) < 0 \Rightarrow$ $[(x + 2) < 0 \land (x + 5) > 0] \lor [(x + 2) > 0 \land (x + 5) < 0]$ $case 1.1 . :$ $x < - 2 \land x > - 5 \Rightarrow$ $x \in] - 5; - 2 [$ $case 1.2 . :$ $x > - 2 \land x < - 5 \Rightarrow no solution$ $case 2 . :$ $x ⩾ 0$ $x^{2} - 7 x + 10 < 0$ $(x - 2) (x - 5) < 0 \Rightarrow$ $[(x - 2) < 0 \land (x - 5) > 0] \lor [(x - 2) > 0 \land (x - 5) < 0]$ $case 2.1 . :$ $x < 2 \land x > 5 \Rightarrow no solution$ $case 2.2 . :$ $x > 2 \land x < 5 \Rightarrow$ $x \in] 2; 5 [$ $at x = 3 the limit of the function is - \infty < 0$ $solution is x \in] - 5; - 2 [\cap] 2; 5 [or$ $∣ x ∣\in] 2; 5 [or 2 <∣ x ∣< 5$
	Commented by NECx last updated on 24/Apr/18

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