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Question Number 84448 by jagoll last updated on 13/Mar/20

5^((x^2 −7∣x∣+10)/(x^2 −6x+9)) < 1

$$\mathrm{5}^{\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{7}\mid\mathrm{x}\mid+\mathrm{10}}{\mathrm{x}^{\mathrm{2}} −\mathrm{6x}+\mathrm{9}}} \:<\:\mathrm{1} \\ $$

Answered by john santu last updated on 13/Mar/20

((x^2 −7∣x∣+10)/(x^2 −6x+9)) < 0 (1) x^2 −6x+9 ≥ 0 ∀xR ⇒ x^2 −7∣x∣+10 < 0 x ≥0 ⇒ (x−5)(x−2) < 0 , 2<x<5 x < 0 ⇒ (x+2)(x+5) < 0, −5<x<−2 solution −5<x<−2 ∨ 2<x<5 ∧ x≠ 3

$$\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{7}\mid\mathrm{x}\mid+\mathrm{10}}{\mathrm{x}^{\mathrm{2}} −\mathrm{6x}+\mathrm{9}}\:<\:\mathrm{0} \\ $$ $$\left(\mathrm{1}\right)\:\mathrm{x}^{\mathrm{2}} −\mathrm{6x}+\mathrm{9}\:\geqslant\:\mathrm{0}\:\forall\mathrm{x}\mathbb{R} \\ $$ $$\Rightarrow\:\mathrm{x}^{\mathrm{2}} −\mathrm{7}\mid\mathrm{x}\mid+\mathrm{10}\:<\:\mathrm{0} \\ $$ $$\mathrm{x}\:\geqslant\mathrm{0}\:\Rightarrow\:\left(\mathrm{x}−\mathrm{5}\right)\left(\mathrm{x}−\mathrm{2}\right)\:<\:\mathrm{0}\:,\:\mathrm{2}<\mathrm{x}<\mathrm{5} \\ $$ $$\mathrm{x}\:<\:\mathrm{0}\:\Rightarrow\:\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{x}+\mathrm{5}\right)\:<\:\mathrm{0},\:−\mathrm{5}<\mathrm{x}<−\mathrm{2} \\ $$ $$\mathrm{solution} \\ $$ $$−\mathrm{5}<\mathrm{x}<−\mathrm{2}\:\vee\:\mathrm{2}<\mathrm{x}<\mathrm{5}\:\wedge\:\mathrm{x}\neq\:\mathrm{3} \\ $$

Commented byjohn santu last updated on 13/Mar/20

l yes sir.i′m forgot

$$\mathrm{l}\:\mathrm{yes}\:\mathrm{sir}.\mathrm{i}'\mathrm{m}\:\mathrm{forgot} \\ $$

Commented byjagoll last updated on 13/Mar/20

thank you mister

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{mister} \\ $$

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