Menu Close

Category: Algebra

find-min-x-y-xy-4-4-2-xy-s-t-x-gt-0-gt-y-

Question Number 194321 by CrispyXYZ last updated on 04/Jul/23 $$\mathrm{find}\:\mathrm{min}\:\:−\frac{\left({x}−{y}\right)\left(\frac{{xy}}{\mathrm{4}}−\mathrm{4}\right)^{\mathrm{2}} }{{xy}} \\ $$$$\mathrm{s}.\:\mathrm{t}.\:\:{x}>\mathrm{0}>{y} \\ $$ Answered by Frix last updated on 04/Jul/23 $${f}\left({x},{y}\right)=−\frac{\left({x}−{y}\right)\left(\frac{{xy}}{\mathrm{4}}−\mathrm{4}\right)^{\mathrm{2}} }{{xy}}=−\frac{\left({x}−{y}\right)\left({xy}−\mathrm{16}\right)^{\mathrm{2}} }{\mathrm{16}{xy}}…

if-x-R-amp-x-x-6-2-2-x-

Question Number 194316 by SajaRashki last updated on 03/Jul/23 $${if}\:\:{x}\in{R}\:\:\&\:\:{x}^{{x}^{\mathrm{6}} } =\left(\sqrt{\mathrm{2}}\right)^{\sqrt{\mathrm{2}}} \:\Rightarrow\:\:{x}=? \\ $$ Commented by Frix last updated on 03/Jul/23 $${x}=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$\Rightarrow\:{x}^{\mathrm{6}}…

Question-194295

Question Number 194295 by Abdullahrussell last updated on 02/Jul/23 Answered by BaliramKumar last updated on 02/Jul/23 $$\mathrm{1}!×\mathrm{2}!×\mathrm{3}!×………………×\mathrm{2023}! \\ $$$$\mathrm{1}^{\mathrm{2023}} ×\mathrm{2}^{\mathrm{2022}} ×\mathrm{3}^{\mathrm{2021}} ×………………×\mathrm{2023}^{\mathrm{1}} \\ $$$$\mathrm{1}^{\mathrm{2023}} ×…×\mathrm{5}^{\mathrm{2019}}…

Let-a-b-c-be-positive-real-numbers-prove-that-a-b-b-c-c-a-3-3-abc-a-b-c-4-

Question Number 194163 by York12 last updated on 29/Jun/23 $$\boldsymbol{{Let}}\:\boldsymbol{{a}}\:,\:\boldsymbol{{b}}\:,\:\boldsymbol{{c}}\:\:\:\boldsymbol{{be}}\:\boldsymbol{{positive}}\:\boldsymbol{{real}}\:\boldsymbol{{numbers}} \\ $$$$\boldsymbol{{prove}}\:\boldsymbol{{that}} \\ $$$$\frac{\boldsymbol{{a}}}{\boldsymbol{{b}}}+\frac{\boldsymbol{{b}}}{\boldsymbol{{c}}}+\frac{\boldsymbol{{c}}}{\boldsymbol{{a}}}+\frac{\mathrm{3}^{\mathrm{3}} \sqrt{{abc}}}{\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}}\geqslant\mathrm{4} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

prove-it-n-k-2-n-k-1-n-k-1-1-k-n-1-

Question Number 194140 by MM42 last updated on 28/Jun/23 $${prove}\:{it}\:: \\ $$$$\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}^{\:\mathrm{2}} \geqslant\:\begin{pmatrix}{\:\:\:\:{n}}\\{{k}−\mathrm{1}}\end{pmatrix}\:×\begin{pmatrix}{\:\:{n}}\\{{k}+\mathrm{1}}\end{pmatrix}\:\:\:\:\:;\:\:\:\mathrm{1}\leqslant{k}\leqslant{n}−\mathrm{1} \\ $$$$ \\ $$ Answered by witcher3 last updated on 28/Jun/23 $$\left(\frac{\mathrm{n}!}{\mathrm{k}!.\left(\mathrm{n}−\mathrm{k}\right)!}\right)^{\mathrm{2}}…