Question Number 198862 by sonukgindia last updated on 25/Oct/23 Answered by Frix last updated on 25/Oct/23 $$\begin{vmatrix}{{a}}&{{b}}&{{c}}\\{{b}}&{{c}}&{{a}}\\{{c}}&{{a}}&{{b}}\end{vmatrix}=\mathrm{3}{abc}−\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \right)= \\ $$$$=\left({a}+{b}+{c}\right)\left({ab}+{ac}+{bc}−\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\right)=…
Question Number 198855 by mokys last updated on 25/Oct/23 $${prove}\::\:\underset{{n}\rightarrow\infty} {{lim}}\:\frac{{n}!}{{n}^{{x}} \left({n}−{x}\right)!}\:=\:\mathrm{1}\: \\ $$ Answered by witcher3 last updated on 25/Oct/23 $$\frac{\Gamma\left(\mathrm{n}+\mathrm{1}\right)}{\Gamma\left(\mathrm{n}+\mathrm{1}−\mathrm{x}\right)\mathrm{n}^{\mathrm{x}} } \\ $$$$\Gamma\left(\mathrm{z}\right)=\sqrt{\mathrm{2}\pi}.\mathrm{z}^{\mathrm{z}−\frac{\mathrm{1}}{\mathrm{2}}}…
Question Number 198849 by sonukgindia last updated on 25/Oct/23 Commented by mr W last updated on 25/Oct/23 $${for}\:{a},{b},\:{c}\:\in{R}\:{no}\:{solution}! \\ $$ Answered by AST last updated…
Question Number 198806 by sonukgindia last updated on 24/Oct/23 Answered by Frix last updated on 24/Oct/23 $$\int\frac{\mathrm{7}{x}^{\mathrm{2}} +\mathrm{5}}{{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{25}}{dx}= \\ $$$$=\int\left(\frac{\mathrm{3}{x}+\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{5}\right)}−\frac{\mathrm{3}{x}−\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}\right)}\right){dx}= \\…
Question Number 198785 by sonukgindia last updated on 24/Oct/23 Answered by AST last updated on 24/Oct/23 $$\frac{\mathrm{1}}{\mathrm{197}}\equiv{x}\left({mod}\:\mathrm{3000}\right)\Rightarrow\mathrm{197}{x}\equiv\mathrm{1}\left({mod}\:\mathrm{3000}\right) \\ $$$$\mathrm{197}{x}\equiv\mathrm{1}\left({mod}\:\mathrm{3}\right)\Rightarrow−{x}\overset{\mathrm{3}} {\equiv}\mathrm{1}\Rightarrow{x}\overset{\mathrm{3}} {\equiv}\mathrm{2} \\ $$$$\mathrm{197}{x}\equiv\mathrm{1}\left({mod}\:\mathrm{8}\right)\Rightarrow\mathrm{5}{x}\equiv\mathrm{1}\left({mod}\:\mathrm{8}\right)\Rightarrow{x}\overset{\mathrm{8}} {\equiv}\mathrm{5} \\…
Question Number 198786 by sonukgindia last updated on 24/Oct/23 Commented by Frix last updated on 24/Oct/23 $$\mathrm{If}\:{a},\:{b},\:{c},\:{d},\:{e}\:\in\mathbb{N}: \\ $$$$\mathrm{1}.\:{abc}=\mathrm{2}^{\mathrm{3}} ×\mathrm{3} \\ $$$$\mathrm{2}.\:{bcd}=\mathrm{2}^{\mathrm{2}} ×\mathrm{3}×\mathrm{5} \\ $$$$\mathrm{3}.\:{cde}=\mathrm{2}^{\mathrm{3}}…
Question Number 198565 by tri26112004 last updated on 22/Oct/23 $${Let}\:{a},{b},{c}\:\in\:{R}^{+} \:{such}\:{that}\:{a}+{b}+{c}=\mathrm{1}. \\ $$$$\:{Prove}\:{that}:\:\:\:\:{b}+{c}\geqslant\mathrm{16}{abc} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 198624 by Tinku Tara last updated on 22/Oct/23 $$\mathrm{Post}\:\mathrm{have}\:\mathrm{been}\:\mathrm{cleanup}\:\mathrm{for}\:\mathrm{users} \\ $$$$\mathrm{and}\:\mathrm{access}\:\mathrm{blocked} \\ $$ Commented by mr W last updated on 22/Oct/23 $${thanks}\:{alot}\:{sir}!\:{well}\:{done}! \\…
Question Number 198516 by mr W last updated on 21/Oct/23 $${To}\:{administrator}\:{Tinku}\:{Tara}\:{sir}: \\ $$$${please}\:{have}\:{a}\:{look}\:{at}\:{what}\:{the}\: \\ $$$${members}\:\underline{{Hridiana}}\:{and}\:\underline{{HomeAlone}} \\ $$$${are}\:{doing}\:{in}\:{the}\:{forum}! \\ $$$${They}\:{are}\:{rioting}! \\ $$$${Please}\:{ban}\:{them}\:{from}\:{the}\:{forum}! \\ $$ Commented by…
Question Number 198489 by sonukgindia last updated on 21/Oct/23 Answered by cortano12 last updated on 21/Oct/23 $$\:\:\mathrm{x}^{\mathrm{16}\left(\mathrm{log}\:_{\mathrm{5}} \mathrm{x}\right)^{\mathrm{3}} −\mathrm{68log}\:_{\mathrm{5}} \mathrm{x}\:} =\:\mathrm{5}^{−\mathrm{16}} \\ $$$$\:\:\mathrm{let}\:\mathrm{log}\:_{\mathrm{5}} \mathrm{x}\:=\:\mathrm{t} \\…