Question Number 30522 by abdo imad last updated on 22/Feb/18 $${let}\:{p}\left({x}\right)=\:\left(\mathrm{1}+{ix}\right)^{{n}} \:−\left(\mathrm{1}−{ix}\right)^{{n}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{roots}\:{of}\:{p}\left({x}\right)\:{and}\:{factorize}\:{p}\left({x}\right) \\ $$$$\left.\right)\:{give}\:{p}\left({x}\right)\:{at}\:{form}\:{of}\:{arcs}. \\ $$$$ \\ $$ Commented by abdo imad last…
Question Number 161590 by geron last updated on 20/Dec/21 $${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−\sqrt{{x}}}\:+\:\frac{\mathrm{1}}{\mathrm{1}−\sqrt{\mathrm{1}−\boldsymbol{{x}}}} \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)_{\boldsymbol{{min}}} =\:? \\ $$ Answered by cortano last updated on 20/Dec/21 Answered by mr…
Question Number 30513 by Penguin last updated on 22/Feb/18 Commented by Penguin last updated on 22/Feb/18 $${f}\left({x}\right)=\mathrm{sin}\left(\lfloor{x}\rfloor\right)\left(−\mathrm{1}\right)^{\lfloor{x}\rfloor} \\ $$$$\mathrm{or},\:\boldsymbol{{similarly}}: \\ $$$${f}\left({x}\right)=\mathrm{sin}\left({x}\right)\left(−\mathrm{1}\right)^{{x}} ,\:{x}\in\mathbb{Z}\: \\ $$$$\: \\…
Question Number 30501 by abdo imad last updated on 22/Feb/18 $${let}\:{put}\:{w}={e}^{{i}\frac{\mathrm{2}\pi}{{n}}} \:\:\:\:\:{find}\:\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\left({x}+{w}^{{k}} \right)^{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\:\sum_{{k}=\mathrm{1}} ^{{n}} {n}\left({x}+{w}^{{k}} \right)^{{n}−\mathrm{1}} \:\:. \\ $$$$ \\ $$…
Question Number 161564 by HongKing last updated on 19/Dec/21 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{values}\:\:\mathrm{x};\mathrm{y};\mathrm{z}>\mathrm{0}\:\:\mathrm{such}\:\mathrm{that}: \\ $$$$\begin{cases}{\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{2z}\:=\:\mathrm{6}}\\{\frac{\mathrm{3}}{\mathrm{y}}\centerdot\left(\frac{\mathrm{2}}{\mathrm{x}}\:+\:\frac{\mathrm{1}}{\mathrm{y}}\right)\:=\:\mathrm{4}\centerdot\left(\frac{\mathrm{2}}{\mathrm{x}\:+\:\mathrm{y}}\:+\:\frac{\mathrm{1}}{\mathrm{2y}}\right)^{\mathrm{2}} }\\{\mathrm{x}\:+\:\mathrm{2}^{\boldsymbol{\mathrm{y}}} \:+\:\mathrm{log}_{\mathrm{2}} \boldsymbol{\mathrm{z}}\:=\:\mathrm{4}}\end{cases} \\ $$ Commented by Rasheed.Sindhi last updated on 20/Dec/21 $$\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left(\mathrm{1},\mathrm{1},\mathrm{2}\right)…
Question Number 161567 by HongKing last updated on 19/Dec/21 $$\mathrm{let}\:\:\mathrm{f}\left(\mathrm{x}\right)\:\:\mathrm{be} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{x}}{\mathrm{ln}\left(\mathrm{1}\:-\:\mathrm{x}\right)} \\ $$$$\mathrm{prove}\:\mathrm{there}\:\mathrm{exists}\:\mathrm{a}\:\mathrm{sequence}\:\left\{\mathrm{a}_{\boldsymbol{\mathrm{k}}} \right\}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{D}\left[\mathrm{f}\left(\mathrm{x}\right)\right]\:=\:\left[\underset{\mathrm{0}} {\overset{\:\infty} {\sum}}\:\mathrm{a}_{\boldsymbol{\mathrm{k}}} \:\mathrm{x}^{\boldsymbol{\mathrm{k}}} \right]\:\left[\mathrm{f}\left(\mathrm{x}\right)\right]^{\mathrm{2}} \\ $$ Answered by…
Question Number 161566 by HongKing last updated on 19/Dec/21 $$\mathrm{if}\:\:\mathrm{x};\mathrm{y};\mathrm{z}>\mathrm{0}\:\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\Sigma\:\frac{\mathrm{x}}{\:\sqrt[{\mathrm{3}}]{\mathrm{y}^{\mathrm{3}} \:+\:\mathrm{25xyz}\:+\:\mathrm{z}^{\mathrm{3}} }}\:\geqslant\:\mathrm{1} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 161563 by HongKing last updated on 19/Dec/21 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\:…}}}\:=\:\mathrm{x}\centerdot\sqrt{\mathrm{x}\centerdot\sqrt{\mathrm{x}\centerdot\sqrt{\mathrm{x}\centerdot\:…}}} \\ $$$$\mathrm{where}\:,\:\mathrm{x}>\mathrm{0} \\ $$ Answered by MJS_new last updated on 20/Dec/21 $$\mathrm{lhs}=\left(\mathrm{lhs}−{x}\right)^{\mathrm{2}} \wedge\mathrm{lhs}\geqslant{x}\:\Rightarrow\:\mathrm{lhs}=\frac{\mathrm{2}{x}+\mathrm{1}+\sqrt{\mathrm{4}{x}+\mathrm{1}}}{\mathrm{2}}…
Question Number 161558 by Rasheed.Sindhi last updated on 19/Dec/21 $${What}'{s}\:{the}\:{value}\:{of}\:{a}\:{for}\:{which} \\ $$$${x}^{\mathrm{2}} +{x}={a}\:\&\:{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}={a}\:{have}\:{one} \\ $$$${root}\:{common}? \\ $$ Answered by MJS_new last updated on 19/Dec/21…
Question Number 96021 by SmNayon11 last updated on 29/May/20 $$\mathrm{if}\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}\:\mathrm{are}\:\mathrm{two}\:\mathrm{complex}\:\mathrm{number} \\ $$$$\mathrm{and}\:\mathrm{p}×\mathrm{q}=\mathrm{m}\:\:,\mathrm{m}\:\mathrm{is}\:\mathrm{a}\:\mathrm{real}\:\mathrm{number}\:. \\ $$$$ \\ $$$$\mathrm{is}\:\mathrm{there}\:\mathrm{always}\:\mathrm{exists}\:\mathrm{a}\:\:\mathrm{p}^{\frac{\mathrm{1}}{\mathrm{3}}} \:\mathrm{and}\:\mathrm{q}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\left(\mathrm{we}\:\mathrm{know}\:\mathrm{p}^{\frac{\mathrm{1}}{\mathrm{3}}} \:\mathrm{and}\:\mathrm{q}^{\frac{\mathrm{1}}{\mathrm{3}}} \mathrm{each}\:\mathrm{has}\:\mathrm{actually}\:\right. \\ $$$$\left.\mathrm{3}\:\mathrm{values}\right) \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{p}^{\frac{\mathrm{1}}{\mathrm{3}}}…