Question Number 74880 by aliesam last updated on 03/Dec/19
$${solve}\:{inR} \\ $$$$\sqrt[{\mathrm{5}}]{\mid{x}+\mathrm{1}\mid}−\sqrt[{\mathrm{10}}]{{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{9}}=\left(\mathrm{2}{x}−\mathrm{10}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$
Answered by mind is power last updated on 03/Dec/19
![x^2 +4x−9 x^2 +4x−9>0⇒ x∈]−∞,((−4−(√(52)))/2)]∪]((−4+(√(52)))/2),+∞[ ((∣x+1∣))^(1/5) ≥((x^2 +4x−9))^(1/(10)) ⇒x^2 +2x+1≥x^2 +4x−9 ⇒2x≤10⇒x≤5....h let α solution of ((∣x+1∣))^(1/5) −((x^2 +4x−9))^(1/(10)) =(2x−10)(√(x^2 +1)) if α<5 we get by..H ((∣α+1∣))^(1/5) −((α^2 +4α−9))^(1/(10)) >0 but (2α−10)(√(α^2 +1))<0 absurd if α>5 ((∣α+1∣))^(1/5) −((α^2 +4α−9))^(1/(10)) <0 (2α−10)(√(α^2 +1))>0 absurd So if ther exist solution α,α∈R−{]−∞,5[∪]5,+∞[}⇔α∈{5} for α=5 we have (6)^(1/5) −((36))^(1/(10)) =(2.5−10)(√(26))=0 true since 36=6^2 S={5}](https://www.tinkutara.com/question/Q74893.png)
$$\mathrm{x}^{\mathrm{2}} +\mathrm{4x}−\mathrm{9} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{4x}−\mathrm{9}>\mathrm{0}\Rightarrow \\ $$$$\left.\mathrm{x}\left.\in\left.\right]−\infty,\frac{−\mathrm{4}−\sqrt{\mathrm{52}}}{\mathrm{2}}\right]\cup\right]\frac{−\mathrm{4}+\sqrt{\mathrm{52}}}{\mathrm{2}},+\infty\left[\right. \\ $$$$\sqrt[{\mathrm{5}}]{\mid\mathrm{x}+\mathrm{1}\mid}\geqslant\sqrt[{\mathrm{10}}]{\mathrm{x}^{\mathrm{2}} +\mathrm{4x}−\mathrm{9}} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{1}\geqslant\mathrm{x}^{\mathrm{2}} +\mathrm{4x}−\mathrm{9} \\ $$$$\Rightarrow\mathrm{2x}\leqslant\mathrm{10}\Rightarrow\mathrm{x}\leqslant\mathrm{5}….\mathrm{h} \\ $$$$\mathrm{let}\:\alpha\:\mathrm{solution} \\ $$$$\mathrm{of}\: \\ $$$$\sqrt[{\mathrm{5}}]{\mid\mathrm{x}+\mathrm{1}\mid}−\sqrt[{\mathrm{10}}]{\mathrm{x}^{\mathrm{2}} +\mathrm{4x}−\mathrm{9}}=\left(\mathrm{2x}−\mathrm{10}\right)\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{if}\:\alpha<\mathrm{5} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{by}..\mathrm{H} \\ $$$$\sqrt[{\mathrm{5}}]{\mid\alpha+\mathrm{1}\mid}−\sqrt[{\mathrm{10}}]{\alpha^{\mathrm{2}} +\mathrm{4}\alpha−\mathrm{9}}\:\:>\mathrm{0} \\ $$$$\mathrm{but}\:\left(\mathrm{2}\alpha−\mathrm{10}\right)\sqrt{\alpha^{\mathrm{2}} +\mathrm{1}}<\mathrm{0}\:\mathrm{absurd} \\ $$$$\mathrm{if}\:\alpha>\mathrm{5} \\ $$$$\sqrt[{\mathrm{5}}]{\mid\alpha+\mathrm{1}\mid}−\sqrt[{\mathrm{10}}]{\alpha^{\mathrm{2}} +\mathrm{4}\alpha−\mathrm{9}}<\mathrm{0} \\ $$$$\left(\mathrm{2}\alpha−\mathrm{10}\right)\sqrt{\alpha^{\mathrm{2}} +\mathrm{1}}>\mathrm{0}\:\mathrm{absurd} \\ $$$$\mathrm{So}\:\mathrm{if}\:\mathrm{ther}\:\mathrm{exist}\:\mathrm{solution}\:\alpha,\alpha\in\mathbb{R}−\left\{\right]−\infty,\mathrm{5}\left[\cup\right]\mathrm{5},+\infty\left[\right\}\Leftrightarrow\alpha\in\left\{\mathrm{5}\right\} \\ $$$$\mathrm{for}\:\alpha=\mathrm{5}\:\mathrm{we}\:\mathrm{have} \\ $$$$\sqrt[{\mathrm{5}}]{\mathrm{6}}−\sqrt[{\mathrm{10}}]{\mathrm{36}}=\left(\mathrm{2}.\mathrm{5}−\mathrm{10}\right)\sqrt{\mathrm{26}}=\mathrm{0} \\ $$$$\mathrm{true}\:\mathrm{since}\:\mathrm{36}=\mathrm{6}^{\mathrm{2}} \\ $$$$\mathrm{S}=\left\{\mathrm{5}\right\} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$