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

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

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

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

Commented byjohn santu last updated on 13/Mar/20

$$\mathrm{l}\mathrm{yes}\mathrm{sir}.{\mathrm{i}}^{\prime }\mathrm{m}\mathrm{forgot}$$

Commented byjagoll last updated on 13/Mar/20

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

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