Question Number 74786 by chess1 last updated on 30/Nov/19 Commented by abdomathmax last updated on 30/Nov/19 $${S}\:=\sum_{{k}=\mathrm{1}} ^{\mathrm{2019}} \:{k}\left(\frac{\mathrm{1}}{\mathrm{2020}}\right)^{{k}} \:={w}\left(\frac{\mathrm{1}}{\mathrm{2020}}\right)\:{with} \\ $$$${w}\left({x}\right)=\sum_{{k}=\mathrm{1}} ^{\mathrm{2019}} \:{kx}^{{k}} \:\:\:{we}\:{have}\:{for}\:{x}\neq\mathrm{1}…
Question Number 9236 by tawakalitu last updated on 24/Nov/16 $$\mathrm{Solve}\:\:\mathrm{simultaneously} \\ $$$$\frac{\mathrm{x}}{\mathrm{y}\:+\:\mathrm{1}_{\:} }\:+\:\frac{\mathrm{y}}{\mathrm{x}\:+\:\mathrm{1}}\:=\:\frac{\mathrm{5}}{\mathrm{3}}\:\:\:\:………….\:\left(\mathrm{i}\right) \\ $$$$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{2}\:\:\:\:\:\:\:\:………..\:\left(\mathrm{ii}\right) \\ $$ Answered by RasheedSoomro last updated on…
Question Number 9229 by tawakalitu last updated on 24/Nov/16 $$\mathrm{2}^{\mathrm{3x}\:+\:\mathrm{1}} \:−\:\mathrm{3}.\mathrm{2}^{\mathrm{2x}} \:+\:\mathrm{2}^{\mathrm{x}\:+\:\mathrm{1}} \:=\:\mathrm{2x} \\ $$$$\mathrm{Find}\:\mathrm{x} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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…