Question Number 164155 by HongKing last updated on 14/Jan/22 $$\mathrm{if}\:\mathrm{the}\:\mathrm{root}\:\mathrm{of}\:\mathrm{equation} \\ $$$$\mathrm{x}^{\mathrm{9}} \:-\:\mathrm{10x}\:+\:\mathrm{1}\:=\:\mathrm{0}\:\:\mathrm{are}\:\:\mathrm{x}_{\mathrm{1}} \:\:\mathrm{and}\:\:\mathrm{x}_{\mathrm{2}} \\ $$$$\mathrm{x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} \:=\:\mathrm{1} \\ $$$$\mathrm{find}:\:\:\mathrm{x}_{\mathrm{1}} \:+\:\mathrm{x}_{\mathrm{2}} \:=\:? \\ $$ Commented…
Question Number 164148 by mathlove last updated on 14/Jan/22 $${ln}\left(\frac{\mathrm{3}^{{x}} }{\frac{\mathrm{27}}{\mathrm{81}}}\right)=\mathrm{0}\:\:\:\:\:\:\:\:\:{then}\:\:\:\:{x}=? \\ $$ Answered by nikif99 last updated on 14/Jan/22 $$\mathrm{log}\:_{{b}} \mathrm{1}=\mathrm{0}\:\Rightarrow \\ $$$$\left(\frac{\mathrm{3}^{{x}} }{\frac{\mathrm{27}}{\mathrm{81}}}\right)=\mathrm{1}\:\Rightarrow\mathrm{3}^{{x}}…
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Question Number 164135 by mathlove last updated on 14/Jan/22 $${p}\left({x}\right)+{xp}\left(−{x}\right)={x}^{\mathrm{2}} +\mathrm{1} \\ $$$${faind}\:\:\:{p}\left(\mathrm{2}\right)=? \\ $$ Answered by mr W last updated on 14/Jan/22 $${p}\left({x}\right)+{xp}\left(−{x}\right)={x}^{\mathrm{2}} +\mathrm{1}\:\:\:…\left({i}\right)…
Question Number 164125 by HongKing last updated on 14/Jan/22 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\underset{\:\mathrm{0}} {\overset{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} {\int}}\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\mathrm{arctan}\:\left(\frac{\mathrm{sin}\boldsymbol{\mathrm{x}}}{\mathrm{u}\:+\:\mathrm{cos}\boldsymbol{\mathrm{x}}}\right)\:\mathrm{dudx}\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{16}}\:+\:\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{ln}\left(\mathrm{2}\right)\:-\:\frac{\pi}{\mathrm{4}} \\ $$ Answered by Kamel last updated on…
Question Number 164124 by HongKing last updated on 14/Jan/22 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\frac{\mathrm{16}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{4}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{1}^{\boldsymbol{\mathrm{x}}} }{\mathrm{4}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{1}^{\boldsymbol{\mathrm{x}}} }\:=\:\frac{\mathrm{8}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{1}}{\mathrm{65}} \\ $$ Answered by MJS_new last…
Question Number 164116 by mathlove last updated on 14/Jan/22 $$\left({a}+{b}+{c}+{x}\right)^{\mathrm{100}} \\ $$$$\mathrm{50}^{{th}} \:{limit}\:{is}\:{equl}\:{to}=? \\ $$ Commented by mr W last updated on 14/Jan/22 $${what}\:{is}\:{a}\:{first}\:{limit}? \\…
Question Number 98570 by HamraboyevFarruxjon last updated on 14/Jun/20 $$\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}}>\mathrm{0}\:\:\:\:\:\:\:\boldsymbol{{prove}}: \\ $$$$\frac{\boldsymbol{{a}}}{\:\sqrt{\boldsymbol{{a}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{{bc}}}}+\frac{\boldsymbol{{b}}}{\:\sqrt{\boldsymbol{{b}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{{ac}}}}+\frac{\boldsymbol{{c}}}{\:\sqrt{\boldsymbol{{c}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{{ab}}}}\geqslant\mathrm{1} \\ $$$$\boldsymbol{{help}}\:\boldsymbol{{please}}… \\ $$ Commented by MJS last updated on…
Question Number 33032 by rahul 19 last updated on 09/Apr/18 $${f}:{N}\rightarrow{R} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{2005}. \\ $$$${and}\: \\ $$$${f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)+……+{f}\left({n}\right)=\:{n}^{\mathrm{2}} \:{f}\left({n}\right),{n}>\mathrm{1}. \\ $$$${Then}\:{f}\left(\mathrm{2004}\right)=? \\ $$ Commented by Joel578…
Question Number 164077 by mathlove last updated on 13/Jan/22 Answered by cortano1 last updated on 13/Jan/22 $$\:\left(\mathrm{1}\right){x}+{y}+{xy}+\mathrm{1}=\mathrm{4}\Rightarrow\left({x}+\mathrm{1}\right)\left({y}+\mathrm{1}\right)=\mathrm{4} \\ $$$$\left(\mathrm{2}\right){y}+{z}+{yz}+\mathrm{1}=\mathrm{6}\Rightarrow\left({y}+\mathrm{1}\right)\left({z}+\mathrm{1}\right)=\mathrm{6} \\ $$$$\left(\mathrm{3}\right)\:{z}+{x}+{zx}+\mathrm{1}=\mathrm{8}\Rightarrow\left({z}+\mathrm{1}\right)\left({x}+\mathrm{1}\right)=\mathrm{8} \\ $$$$\left(\mathrm{1}\right)×\left(\mathrm{2}\right)×\left(\mathrm{3}\right)\Rightarrow\left({x}+\mathrm{1}\right)\left({y}+\mathrm{1}\right)\left({z}+\mathrm{1}\right)=\pm\sqrt{\mathrm{4}×\mathrm{4}×\mathrm{4}×\mathrm{3}}=\pm\mathrm{8}\sqrt{\mathrm{3}} \\ $$$$\:{z}+\mathrm{1}=\pm\mathrm{2}\sqrt{\mathrm{3}}…