Question Number 161295 by Ari last updated on 15/Dec/21 $${prove}\:{that}:{x}^{\mathrm{8}} +{x}^{\mathrm{6}} −{x}^{\mathrm{3}} −{x}+\mathrm{1}>\mathrm{0},{x}\in{R} \\ $$ Commented by Ari last updated on 15/Dec/21 thanks Mr! Commented by…
Question Number 30218 by abdo imad last updated on 18/Feb/18 $${prove}\:{that}\:{D}\left({x}^{\mathrm{5}} −\mathrm{1},{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)=\mathrm{1}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 30220 by abdo imad last updated on 18/Feb/18 $${let}\:{p}\left({x}\right)=\:{x}^{\mathrm{3}} \:{px}\:+{q} \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:{p}\:{have}\:{double}\:{roots}\Leftrightarrow\:\mathrm{4}{p}^{\mathrm{3}} \:+\mathrm{27}{q}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\:{let}\:{suppose}\:{p}\:{have}\:\mathrm{3}\:{real}\:{roots}\:{differnts}\:{prove}\:{that} \\ $$$$\mathrm{4}{p}^{\mathrm{3}} \:+\mathrm{27}{q}^{\mathrm{2}} \:<\mathrm{0}. \\ $$ Commented…
Question Number 161280 by HongKing last updated on 15/Dec/21 $$\mathrm{if}\:\:\mathrm{x};\mathrm{y};\mathrm{z}>\mathrm{0}\:\:\mathrm{and}\:\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}}\:+\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{y}}\:+\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{z}}\:=\:\mathrm{1} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\:\geqslant\:\frac{\mathrm{3}}{\mathrm{4}}\:\mathrm{xyz} \\ $$ Answered by 1549442205PVT last updated on 16/Dec/21 $${From}\:{the}\:{hypothesis}\:\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}}\:+\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{y}}\:+\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{z}}\:=\:\mathrm{1} \\…
Question Number 95742 by Abdulrahman last updated on 27/May/20 $$\frac{\mathrm{0}}{\mathrm{0}}=\mathrm{2} \\ $$$$\frac{\mathrm{100}−\mathrm{100}}{\mathrm{100}−\mathrm{100}}=\frac{\mathrm{10}^{\mathrm{2}} −\mathrm{10}^{\mathrm{2}} }{\mathrm{10}^{\mathrm{2}} −\mathrm{10}^{\mathrm{2}} }=\frac{\left(\mathrm{10}+\mathrm{10}\right)\left(\mathrm{10}−\mathrm{10}\right)}{\mathrm{10}\left(\mathrm{10}−\mathrm{10}\right)} \\ $$$$\frac{\mathrm{20}}{\mathrm{10}}=\mathrm{2} \\ $$$$\mathrm{where}\:\mathrm{is}\:\mathrm{the}\:\mathrm{mastike} \\ $$ Commented by bobhans…
Question Number 161242 by HongKing last updated on 14/Dec/21 $$\mathrm{x}\:;\:\mathrm{y}\:;\:\mathrm{z}\:<\:\mathrm{0} \\ $$$$\frac{\mathrm{x}^{\mathrm{3}} }{\:\sqrt{\mathrm{yz}}}\:=\:-\mathrm{3}\:\:;\:\:\frac{\mathrm{y}^{\mathrm{3}} }{\:\sqrt{\mathrm{xz}}}\:=\:-\mathrm{6}\:\:;\:\:\frac{\mathrm{z}^{\mathrm{3}} }{\:\sqrt{\mathrm{xy}}}\:=\:-\mathrm{8} \\ $$$$\mathrm{find}\:\:\mathrm{x}\centerdot\mathrm{y}\centerdot\mathrm{z}\:=\:? \\ $$ Answered by mr W last updated…
Question Number 161234 by mathlove last updated on 14/Dec/21 $$\mathrm{1}:!{x}=? \\ $$$$\mathrm{2}:!!{x}=? \\ $$$$\mathrm{3}:{x}!!=?\:\:\:\:{when}\:\:\:{X}\:\:{is}\:\:{odd} \\ $$$$\mathrm{4}:{x}!!=?\:\:\:\:{when}\:\:{X}\:\:{is}\:\:{evan} \\ $$$$\mathrm{5}:{x}!!!=?\:\:\:\:\:\:\:{when}\:{X}\:{is}\:{odd} \\ $$$$\mathrm{6}:\mathrm{x}!!!=?\:\:\:\:\:\:\:{when}\:{X}\:\:{is}\:{evan} \\ $$ Commented by mr…
Question Number 161215 by mathlove last updated on 14/Dec/21 $$\mathrm{log}\:\underset{\mathrm{2}} {{x}}+{log}\underset{{x}} {\mathrm{2}}=\frac{{e}^{\mathrm{2}} +\mathrm{1}}{{e}}\:\:\:\:\:\:\:\:\:\:{x}=? \\ $$ Commented by cortano last updated on 14/Dec/21 $$\:{let}\:\begin{cases}{\mathrm{log}\:_{\mathrm{2}} \left({x}\right)={u}}\\{\mathrm{log}\:_{{x}} \left(\mathrm{2}\right)=\frac{\mathrm{1}}{{u}}}\end{cases}…
Question Number 95668 by I want to learn more last updated on 26/May/20 $$\mathrm{Prove}\:\mathrm{that}:\:\:\:\mathrm{2}^{\mathrm{1}/\mathrm{4}} .\mathrm{4}^{\mathrm{1}/\mathrm{8}} .\mathrm{8}^{\mathrm{1}/\mathrm{16}} .\mathrm{16}^{\mathrm{1}/\mathrm{32}} .\:\:…\:\:\infty\:\:\:=\:\:\:\mathrm{2} \\ $$ Commented by Tony Lin last…
Question Number 161196 by Ar Brandon last updated on 13/Dec/21 $$\mathrm{Montrons}\:\mathrm{que}\:\mathrm{B}_{\mathrm{2}} \left(\mathrm{0},\:\mathrm{1}\right)\:\mathrm{est}\:\mathrm{homeomorphe}\:\grave {\mathrm{a}}\:\mathbb{R}^{\mathrm{2}} . \\ $$$$\mathrm{O}\grave {\mathrm{u}}\:\mathrm{B}_{\mathrm{2}} \left(\mathrm{0},\:\mathrm{1}\right)=\mathrm{boule}\:\mathrm{unit}\acute {\mathrm{e}}\:\mathrm{de}\:\mathbb{R}^{\mathrm{2}} . \\ $$ Terms of Service…