Category: Algebra

Question Number 221238 by fantastic last updated on 28/May/25 $${a}=\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\:{then}\:\left\{\sqrt{{a}}−\frac{\mathrm{1}}{\:\sqrt{{a}}}\right\}=?? \\ $$ Answered by Rasheed.Sindhi last updated on 28/May/25 $${a}=\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\:{then}\:\left\{\sqrt{{a}}−\frac{\mathrm{1}}{\:\sqrt{{a}}}\right\}=?? \\ $$$${a}=\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{3}}\:\right)^{\mathrm{2}} +\mathrm{2}\left(\sqrt{\mathrm{2}}\:\right)\left(\sqrt{\mathrm{3}}\:\right) \\…

Question Number 221262 by Nicholas666 last updated on 28/May/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{if}\:{a},\:{b},\:{c},\:>\:\mathrm{0}\:\:,\:\mathrm{show}\:\mathrm{that}; \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{a}^{\mathrm{5}} }{{b}^{\mathrm{2}} }\:+\:\frac{{b}}{{c}}\:+\:\frac{{c}^{\mathrm{3}} }{{a}^{\mathrm{2}} }\:>\:\mathrm{2}{a} \\ $$$$ \\ $$ Commented by mr…

Question Number 221195 by Rojarani last updated on 26/May/25 Commented by Ghisom last updated on 26/May/25 $$\mathrm{I}\:\mathrm{get}\:\frac{\mathrm{49}}{\mathrm{34}} \\ $$ Answered by mr W last updated…

Question Number 221185 by hardmath last updated on 26/May/25 Answered by Spillover last updated on 27/May/25 Answered by Spillover last updated on 27/May/25 Answered by…

Question Number 221168 by universe last updated on 25/May/25 Answered by Frix last updated on 26/May/25 $$\mathrm{Let}\:{b}={pa}\wedge{p}<\mathrm{0} \\ $$$$\lambda=\mathrm{min}\:\left(\frac{{p}−\mathrm{1}}{{ap}}\sqrt{\mathrm{2}{a}^{\mathrm{4}} {p}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{2}} {p}+\mathrm{1}}\right) \\ $$$$\mathrm{Using}\:\mathrm{partial}\:\mathrm{differenciation}\:\mathrm{we}\:\mathrm{get} \\…