Category: Others

Question Number 202082 by Abduljalal last updated on 19/Dec/23 Answered by AST last updated on 20/Dec/23 $${p}^{\mathrm{2}} −{p}−\mathrm{2}=\mathrm{0}\left({p}={x}^{\mathrm{3}} \right)\Rightarrow{p}=\mathrm{2}\:{or}\:−\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{3}} =\mathrm{2}\:{or}\:{x}^{\mathrm{3}} =−\mathrm{1}\Rightarrow{x}=\sqrt[{\mathrm{3}}]{\mathrm{2}};\sqrt[{\mathrm{3}}]{\mathrm{2}}{e}^{\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} ,\sqrt[{\mathrm{3}}]{\mathrm{2}}{e}^{\frac{\mathrm{4}\pi{i}}{\mathrm{3}}} \\…

Question Number 201917 by York12 last updated on 15/Dec/23 $$ \\ $$ Commented by York12 last updated on 15/Dec/23 $$\mathrm{please}\:\mathrm{help} \\ $$ Commented by York12…

Question Number 201815 by Euclid last updated on 13/Dec/23 Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 201491 by tri26112004 last updated on 07/Dec/23 $$\mathrm{1}.\:{x}^{\mathrm{2}} −\mathrm{1}+\sqrt[{\mathrm{3}}]{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} }=\mathrm{10}{x} \\ $$$$\mathrm{2}.\:\sqrt{{x}−\frac{\mathrm{1}}{{x}}}+\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}={x} \\ $$$$\mathrm{3}.\:\sqrt{\mathrm{2}{x}−\frac{\mathrm{8}}{{x}}}+\mathrm{2}\sqrt{\mathrm{1}−\frac{\mathrm{2}}{{x}}}\geqslant{x} \\ $$$$\mathrm{4}.\:\sqrt{{x}^{\mathrm{2}} +{x}}+\sqrt{{x}+\mathrm{2}}\geqslant\sqrt{\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)} \\ $$$$\mathrm{5}.\:\left(\sqrt{{x}+\mathrm{5}}−\sqrt{{x}−\mathrm{3}}\right)\left(\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{15}}\right)\geqslant\mathrm{8} \\…

Question Number 201308 by Mingma last updated on 04/Dec/23 Commented by aleks041103 last updated on 04/Dec/23 $$−{Q}\Leftrightarrow−{Q} \\ $$$${Q}\Leftrightarrow−\left(−{Q}\right) \\ $$$${P}\Rightarrow{Q}\Leftrightarrow−\left(−{Q}\right) \\ $$$$\therefore{P}\Rightarrow−\left({Q}\right) \\ $$…