Category: Algebra

Question Number 193484 by Frix last updated on 15/Jun/23 $$\mathrm{Nice}\:\mathrm{problem}: \\ $$$$\mathrm{Find}\:\mathrm{8}\:\mathrm{distinctive}\:\mathrm{numbers}\:\in\mathbb{N}\backslash\left\{\mathrm{0}\right\}\:\mathrm{such} \\ $$$$\mathrm{that}\:\mathrm{these}\:\mathrm{are}\:\mathrm{simultaniously}\:\mathrm{true}: \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:{a}+{b}+{c}+{d}\:=\:{e}+{f}+{g}+{h} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \:=\:{e}^{\mathrm{2}} +{f}^{\mathrm{2}} +{g}^{\mathrm{2}} +{h}^{\mathrm{2}}…

Question Number 193410 by cortano12 last updated on 13/Jun/23 $$\:\:\begin{cases}{\mathrm{x}=\sqrt{\mathrm{3}−\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{3}}}}}\\{\mathrm{y}=\sqrt{\mathrm{3}+\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{3}}}}}\end{cases}\: \\ $$$$\:\:\:\:\underbrace{\boldsymbol{{x}}} \\ $$ Answered by aba last updated on 13/Jun/23 $$\mathrm{xy}=\sqrt{\mathrm{9}−\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{3}}\right)}=\sqrt{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}}=\sqrt{\mathrm{3}}−\mathrm{1}\:\wedge\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{6} \\…

Question Number 193363 by MATHEMATICSAM last updated on 11/Jun/23 $${y}\:=\:\mathrm{4}\:×\:\mathrm{10}^{\mathrm{2}{x}} \\ $$$$\mathrm{Express}\:{x}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{y},\:\mathrm{giving}\:\mathrm{an}\:\mathrm{exact} \\ $$$$\mathrm{simplified}\:\mathrm{answer}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{log}\:\mathrm{base}\:\mathrm{10}. \\ $$ Answered by aba last updated on 11/Jun/23 $$\mathrm{y}=\mathrm{4}×\mathrm{10}^{\mathrm{2x}} \:\Rightarrow\:\mathrm{10}^{\mathrm{2x}}…

Question Number 193346 by pascal889 last updated on 11/Jun/23 $$\sqrt{\mathrm{4x}−\mathrm{3}}−\sqrt{\mathrm{2x}−\mathrm{5}}=\mathrm{2} \\ $$$$\mathrm{solve}\:\mathrm{for}\:\mathrm{x} \\ $$ Answered by AST last updated on 11/Jun/23 $${u}−{v}=\mathrm{2};\:{u}^{\mathrm{2}} −\mathrm{2}{v}^{\mathrm{2}} =\mathrm{7} \\…

Question Number 193314 by York12 last updated on 10/Jun/23 $$ \\ $$$$\boldsymbol{{a}}\:,\boldsymbol{{b}},\:\boldsymbol{{c}}\:\:>\:\mathrm{0}\:\&\:\:\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} =\mathrm{3}\:\boldsymbol{{prove}}\:\boldsymbol{{that}}\: \\ $$$$\sqrt[{\mathrm{3}}]{\left(\mathrm{1}+\frac{\mathrm{3}}{\boldsymbol{{ab}}+\boldsymbol{{bc}}+\boldsymbol{{ca}}}\:\right)^{\left(\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} } \:}\leqslant\left(\mathrm{1}+\frac{\boldsymbol{{a}}}{\boldsymbol{{b}}}\right)\left(\mathrm{1}+\frac{\boldsymbol{{b}}}{\boldsymbol{{c}}}\right)\left(\mathrm{1}+\frac{\boldsymbol{{c}}}{\boldsymbol{{a}}}\right) \\ $$ Answered by witcher3 last…