Question Number 152365 by mathdanisur last updated on 27/Aug/21 Answered by ghimisi last updated on 28/Aug/21 $${a}={x}+{y};{b}={y}+{z};{c}={x}+{z} \\ $$$${p}={x}+{y}+{z};{q}={ab}+{bc}+{ac};{r}={abc} \\ $$$$\Leftrightarrow….\Leftrightarrow{p}^{\mathrm{3}} +\mathrm{9}{r}\geqslant\mathrm{4}{pq}\Leftrightarrow{schur} \\ $$ Commented…
Question Number 21294 by Tinkutara last updated on 19/Sep/17 $$\mathrm{Let}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} \:\mathrm{be}\:\mathrm{complex}\:\mathrm{numbers},\:\mathrm{not} \\ $$$$\mathrm{all}\:\mathrm{real},\:\mathrm{such}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \mid\:=\:\mid{z}_{\mathrm{2}} \mid\:=\:\mid{z}_{\mathrm{3}} \mid\:=\:\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{2}\left({z}_{\mathrm{1}} \:+\:{z}_{\mathrm{2}} \:+\:{z}_{\mathrm{3}} \right)\:−\:\mathrm{3}{z}_{\mathrm{1}} {z}_{\mathrm{2}} {z}_{\mathrm{3}}…
Question Number 152364 by mathdanisur last updated on 27/Aug/21 Answered by Kamel last updated on 28/Aug/21 $$ \\ $$$$\Omega\left({a},{b}\right)=\int_{\mathrm{0}} ^{\pi} \frac{{Ln}\left({tan}\left({ax}\right)\right)}{\mathrm{1}−\mathrm{2}{bcos}\left({x}\right)+{b}^{\mathrm{2}} }{dx}\:,\:\mid{b}\mid<\mathrm{1},\:\mathrm{0}<{a}\leqslant\frac{\mathrm{1}}{\mathrm{2}}. \\ $$$${We}\:{have}:\:{Ln}\left({tan}\left({ax}\right)\right)=−\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{+\infty}…
Question Number 21293 by Tinkutara last updated on 19/Sep/17 $$\mathrm{Let}\:{n}\:\mathrm{be}\:\mathrm{an}\:\mathrm{even}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{such} \\ $$$$\mathrm{that}\:\frac{{n}}{\mathrm{2}}\:\mathrm{is}\:\mathrm{odd}\:\mathrm{and}\:\mathrm{let}\:\alpha_{\mathrm{0}} ,\:\alpha_{\mathrm{1}} ,\:….,\:\alpha_{{n}−\mathrm{1}} \:\mathrm{be} \\ $$$$\mathrm{the}\:\mathrm{complex}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{unity}\:\mathrm{of}\:\mathrm{order}\:{n}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({a}\:+\:{b}\alpha_{{k}} ^{\mathrm{2}} \right)\:=\:\left({a}^{\frac{{n}}{\mathrm{2}}} \:+\:{b}^{\frac{{n}}{\mathrm{2}}} \right)^{\mathrm{2}}…
Question Number 152358 by Jonathanwaweh last updated on 27/Aug/21 Commented by JDamian last updated on 27/Aug/21 $$\mathrm{5} \\ $$ Answered by Olaf_Thorendsen last updated on…
Question Number 86825 by M±th+et£s last updated on 31/Mar/20 $${a}^{\mathrm{3}} +\frac{\mathrm{1}}{{a}^{\mathrm{3}} }=\mathrm{18} \\ $$$${a}^{\mathrm{4}} +\frac{\mathrm{1}}{{a}^{\mathrm{4}} }=? \\ $$ Commented by Ar Brandon last updated on…
Question Number 152326 by mathdanisur last updated on 27/Aug/21 $$\int\sqrt{\frac{\mathrm{1}\:+\:\mathrm{sin}\boldsymbol{\mathrm{x}}}{\mathrm{cos}\boldsymbol{\mathrm{x}}}}\:\mathrm{dx}\:=\:? \\ $$ Commented by puissant last updated on 27/Aug/21 $$=\int\sqrt{{secx}+{tanx}}{dx} \\ $$$${Q}\mathrm{151568}\: \\ $$ Answered…
Question Number 21248 by Tinkutara last updated on 17/Sep/17 $$\mathrm{The}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{which} \\ $$$$\mathrm{touches}\:\mathrm{the}\:\mathrm{given}\:\mathrm{circles}\:\mid{z}\:−\:{z}_{\mathrm{1}} \mid\:= \\ $$$$\mid\mathrm{3}\:+\:\mathrm{4}{i}\mid\:\mathrm{and}\:\mid{z}\:−\:{z}_{\mathrm{2}} \mid\:=\:\mid\mathrm{1}\:+\:{i}\sqrt{\mathrm{3}}\mid\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{hyperbola},\:\mathrm{then}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{its} \\ $$$$\mathrm{transverse}\:\mathrm{axis}\:\mathrm{is} \\ $$ Terms of Service…
Question Number 152323 by mathdanisur last updated on 27/Aug/21 $$\mathrm{x}^{\mathrm{2}} \centerdot\mathrm{y}=\frac{\mathrm{1}}{\mathrm{18}}\:\:\mathrm{and}\:\:\mathrm{x}\centerdot\mathrm{y}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\mathrm{find}\:\:\left(\mathrm{xy}\right)^{−\mathrm{2}} \:=\:? \\ $$ Answered by Olaf_Thorendsen last updated on 27/Aug/21 $${x}^{\mathrm{2}}…
Question Number 86779 by M±th+et£s last updated on 31/Mar/20 $${ssolve} \\ $$$$\left.\mathrm{1}\right){x}−\left[{x}\right]\geqslant\mathrm{0} \\ $$$$\left.\mathrm{2}\right){x}−\left[{x}\right]\leqslant\mathrm{0} \\ $$$$\left.\mathrm{3}\right){x}+\left[{x}\right]\geqslant\mathrm{0} \\ $$$$\left.\mathrm{4}\right){x}+\left[{x}\right]\leqslant\mathrm{0}\: \\ $$ Answered by mr W last…