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Question Number 184573 by mnjuly1970 last updated on 08/Jan/23 Commented by SEKRET last updated on 08/Jan/23 $$\:\:\:\boldsymbol{\mathrm{use}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{triangle}} \\ $$ Commented by Frix last updated on…
Question Number 184567 by a.lgnaoui last updated on 08/Jan/23 $${Resoudre}\:{dans}\:\mathbb{Z}^{+} \\ $$$$\sqrt{{a}}\:\:+\sqrt{{b}}\:={z}\:\:\:\:\left({a},{b},{z}\right)\in\mathbb{N}^{\mathrm{3}} \\ $$$${a},{b}\:? \\ $$ Answered by Frix last updated on 08/Jan/23 $${a}={n}_{\mathrm{1}} ^{\mathrm{2}}…
Question Number 53491 by ajfour last updated on 22/Jan/19 $$\left(\frac{\mathrm{1}+{x}}{\:\sqrt{{x}}}\right)^{\mathrm{2}} +\mathrm{2}{a}\left(\frac{\mathrm{1}+{x}}{\:\sqrt{{x}}}\right)+\mathrm{1}=\mathrm{0} \\ $$$${solve}\:{for}\:{x}. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19 $${k}^{\mathrm{2}} +\mathrm{2}{ak}+{a}^{\mathrm{2}} +\mathrm{1}={a}^{\mathrm{2}}…
Question Number 184555 by Shrinava last updated on 08/Jan/23 $$\mathrm{Suppose}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\:\mathrm{of} \\ $$$$\mathrm{complex}\:\mathrm{numbers}\:\:\boldsymbol{\mathrm{x}}\:\mathrm{or}\:\boldsymbol{\mathrm{y}}\:\mathrm{is}\:\mathrm{7}\:,\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{their}\:\mathrm{cubes}\:\mathrm{is}\:\mathrm{10}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{largest} \\ $$$$\mathrm{true}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sum}\:\:\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}\:\:\mathrm{that}\:\mathrm{satisfies} \\ $$$$\mathrm{these}\:\mathrm{conditions}. \\ $$$$\left.\mathrm{A}\left.\right)\left.\mathrm{4}\left.\:\left.\:\:\mathrm{B}\right)\mathrm{5}\:\:\:\mathrm{C}\right)\mathrm{6}\:\:\:\mathrm{D}\right)\mathrm{7}\:\:\:\mathrm{E}\right)\mathrm{8} \\ $$ Commented by mr…
Question Number 184551 by a.lgnaoui last updated on 08/Jan/23 $${Resoudre}\:{dans}\:\mathbb{Z}^{+} \\ $$$${x}+{y}+\sqrt{{xy}}\:\:\:\:=\mathrm{39} \\ $$ Answered by Frix last updated on 08/Jan/23 $${xy}\geqslant\mathrm{0}\:\Rightarrow\:{x}\geqslant\mathrm{0}\wedge{y}\geqslant\mathrm{0}\vee{x}<\mathrm{0}\wedge{y}<\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:{x}<\mathrm{0}\wedge{y}<\mathrm{0} \\…
Question Number 184535 by CrispyXYZ last updated on 08/Jan/23 $$\mathrm{If}\:{a},\:{b}>\mathrm{0}\:\mathrm{such}\:\mathrm{that}\:\mathrm{2}{a}+{b}=\mathrm{2}, \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}: \\ $$$$\left.\mathrm{1}\right)\:\:\left(\mathrm{4}{a}^{\mathrm{2}} +\mathrm{1}\right)\left({b}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right)\:\:\frac{\mathrm{2}{a}^{\mathrm{2}} −{b}+\mathrm{4}}{{a}+\mathrm{1}}+\frac{{b}^{\mathrm{2}} −\mathrm{2}{a}−\mathrm{2}}{{b}+\mathrm{4}} \\ $$ Answered by cortano1…
Question Number 118967 by obaidullah last updated on 21/Oct/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\left(\frac{{x}!}{{x}^{{x}} }\right)^{\frac{\mathrm{1}}{{x}}} \right)^{\frac{\mathrm{2}\centerdot\mathrm{4}}{\mathrm{1}\centerdot\mathrm{3}}\centerdot\frac{\mathrm{4}\centerdot\mathrm{6}}{\mathrm{3}\centerdot\mathrm{5}}\centerdot\centerdot\centerdot} =? \\ $$$$\frac{\mathrm{2}}{\mathrm{1}}\centerdot\frac{\mathrm{2}}{\mathrm{3}}\centerdot\frac{\mathrm{4}}{\mathrm{3}}\centerdot\frac{\mathrm{4}}{\mathrm{5}}\centerdot\frac{\mathrm{6}}{\mathrm{5}}\centerdot\frac{\mathrm{6}}{\mathrm{7}}\centerdot\centerdot\centerdot=\frac{\pi}{\mathrm{2}} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\left(\frac{{x}!}{{x}^{{x}} }\right)^{\frac{\mathrm{1}}{{x}}} \right)^{\frac{\mathrm{2}}{\mathrm{1}}\centerdot\frac{\mathrm{4}}{\mathrm{3}}\centerdot\frac{\mathrm{4}}{\mathrm{3}}\centerdot\frac{\mathrm{6}}{\mathrm{5}}\centerdot\centerdot\centerdot} =\left(\frac{\mathrm{1}}{{e}}\right)^{\frac{\pi}{\mathrm{2}}} =\sqrt{\left(\frac{\mathrm{1}}{{e}}\right)^{\pi} }=\frac{\mathrm{1}}{\:\sqrt{{e}^{\pi} }}…
Question Number 184503 by mr W last updated on 07/Jan/23 Answered by SEKRET last updated on 07/Jan/23 $$\:\:\boldsymbol{\mathrm{a}}\:=\sqrt{\mathrm{11}+\boldsymbol{\mathrm{a}}} \\ $$$$\:\:\:\boldsymbol{\mathrm{a}}^{\mathrm{2}} =\:\mathrm{11}\:+\:\boldsymbol{\mathrm{a}} \\ $$$$\:\:\:\boldsymbol{\mathrm{a}}^{\mathrm{3}} =\:\mathrm{11}\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{a}}^{\mathrm{2}} =\mathrm{11}\boldsymbol{\mathrm{a}}+\mathrm{11}+\boldsymbol{\mathrm{a}}=\mathrm{12}\boldsymbol{\mathrm{a}}+\mathrm{11}…
Question Number 184477 by Shrinava last updated on 07/Jan/23 Commented by JDamian last updated on 07/Jan/23 I don't even understand your question. Would you mind express in math language? Commented by mr W last updated on 07/Jan/23…