Question Number 9219 by tawakalitu last updated on 23/Nov/16 $$\mathrm{Show}\:\mathrm{that}:\: \\ $$$$\left(\mathrm{a}\:+\:\mathrm{b}\right)\left[\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{2}} }\right]\left[\frac{\mathrm{a}^{\mathrm{4}} }{\mathrm{b}^{\mathrm{2}} }\:+\:\frac{\mathrm{b}^{\mathrm{4}} }{\mathrm{a}^{\mathrm{2}} }\right]\:\geqslant\:\mathrm{8}\sqrt{\mathrm{ab}} \\ $$ Commented by Yozzias last updated…
Question Number 140291 by ajfour last updated on 06/May/21 Commented by ajfour last updated on 06/May/21 $$\:{S}=\:{a}+\frac{{a}}{\mathrm{2}}+\frac{{a}}{\mathrm{4}}+\frac{{a}}{\mathrm{8}}+……. \\ $$$$\:\:\:=\:\mathrm{2}{a}\:. \\ $$ Commented by Dwaipayan Shikari…
Question Number 74742 by TawaTawa last updated on 30/Nov/19 $$\mathrm{If}\:\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\:\:\:\:\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{x}\:+\:\mathrm{1}\:\:\:=\:\:\mathrm{0},\:\: \\ $$$$\mathrm{Find}\:\:\:\:\:\:\:\:\:\alpha^{\mathrm{23}} \:+\:\beta^{\mathrm{23}} \:\:\:\:\:\mathrm{without}\:\mathrm{demoivre}'\mathrm{s}\:\mathrm{theorem}. \\ $$ Commented by abdomathmax last updated on 02/Dec/19 $${x}^{\mathrm{2}}…
Question Number 140273 by liberty last updated on 06/May/21 $$\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}\:+\mathrm{5x}\:=\:\sqrt{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4}\right)\left(\mathrm{x}+\mathrm{24}\right)} \\ $$ Answered by MJS_new last updated on 06/May/21 $${x}=\mathrm{5} \\ $$$$\mathrm{I}\:\mathrm{saw}\:\mathrm{that}\:\left(\mathrm{5}^{\mathrm{2}} +\mathrm{4}\right)\left(\mathrm{5}+\mathrm{24}\right)=\mathrm{29}^{\mathrm{2}} \\…
Question Number 9199 by tawakalitu last updated on 22/Nov/16 $$\mathrm{If}\:\:\mathrm{r}^{\mathrm{2}} \:=\:\left(\mathrm{x}\:+\:\mathrm{ea}\right)^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{s}^{\mathrm{2}} \:=\:\left(\mathrm{x}\:−\:\mathrm{ea}\right)^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \\ $$$$\mathrm{and}\:\mathrm{r}\:+\:\mathrm{s}\:=\:\mathrm{2a},\:\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{r}\:=\:\mathrm{a}\:+\:\mathrm{ex},\:\:\mathrm{s}\:=\:\mathrm{a}\:−\:\mathrm{ex},\:\mathrm{and}\:\mathrm{that}, \\ $$$$\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}\:−\:\mathrm{e}^{\mathrm{2}} \right)\:+\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{a}^{\mathrm{2}} \left(\mathrm{1}\:−\:\mathrm{e}^{\mathrm{2}}…
Question Number 140275 by liberty last updated on 06/May/21 $$\mathrm{If}\:\begin{cases}{\mathrm{a}+\mathrm{b}+\mathrm{c}\:=\:\mathrm{5}}\\{\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}+\frac{\mathrm{1}}{\mathrm{c}}=\frac{\mathrm{1}}{\mathrm{5}}}\end{cases} \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} \:? \\ $$ Answered by Ar Brandon last updated…
Question Number 140271 by EnterUsername last updated on 06/May/21 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{common}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equations} \\ $$$$\mathrm{x}^{\mathrm{5}} −\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\:\mathrm{and}\:\mathrm{x}^{\mathrm{4}} −\mathrm{1}=\mathrm{0}\:\mathrm{is}\:\_\_\_\_. \\ $$ Answered by liberty last updated on 06/May/21…
Question Number 140260 by EnterUsername last updated on 05/May/21 $$\mathrm{Passage}:\:\mathrm{If}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} \:\mathrm{and}\:{z}_{\mathrm{3}} \:\mathrm{are}\:\mathrm{three}\:\mathrm{complex}\:\mathrm{numbers} \\ $$$$\mathrm{representing}\:\mathrm{the}\:\mathrm{points}\:{A},\:{B}\:\mathrm{and}\:{C},\:\mathrm{respectively},\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{Argands}\:\mathrm{plane}\:\mathrm{and}\:\angle{BAC}=\alpha,\:\mathrm{then} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{z}_{\mathrm{3}} −{z}_{\mathrm{1}} }{{z}_{\mathrm{2}} −{z}_{\mathrm{1}} }=\left(\frac{{AC}}{{AB}}\right)\left(\mathrm{cos}\alpha+{i}\mathrm{sin}\alpha\right) \\ $$$$\left({i}\right)\:\mathrm{If}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}…
Question Number 74716 by chess1 last updated on 29/Nov/19 Answered by Tanmay chaudhury last updated on 29/Nov/19 $${x}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)×{x}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\right)×…{x}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{25}} }\right)=\mathrm{1} \\ $$$${x}^{\mathrm{25}} ×{A}=\mathrm{1} \\ $$$${x}=\left(\frac{\mathrm{1}}{{A}}\right)^{\frac{\mathrm{1}}{\mathrm{25}}}…
Question Number 74712 by aliesam last updated on 29/Nov/19 Commented by MJS last updated on 29/Nov/19 $$\mathrm{easy}\:\mathrm{solutions} \\ $$$${x}={y}=\mathrm{1} \\ $$$${x}={y}=−\mathrm{2} \\ $$$$\mathrm{no}\:\mathrm{other}\:\mathrm{real}\:\mathrm{solutions} \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{2}\right)\:\mathrm{and}\:\mathrm{dividing}\:\mathrm{by}\:{x}^{\mathrm{2}}…