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 $$\mathrm{if}\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}\:\mathrm{are}\:\mathrm{two}\:\mathrm{complex}\:\mathrm{number} \\ $$$$\mathrm{and}\:\mathrm{p}×\mathrm{q}=\mathrm{m}\:\:,\mathrm{m}\:\mathrm{is}\:\mathrm{a}\:\mathrm{real}\:\mathrm{number}\:. \\ $$$$ \\ $$$$\mathrm{is}\:\mathrm{there}\:\mathrm{always}\:\mathrm{exists}\:\mathrm{a}\:\:\mathrm{p}^{\frac{\mathrm{1}}{\mathrm{3}}} \:\mathrm{and}\:\mathrm{q}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\left(\mathrm{we}\:\mathrm{know}\:\mathrm{p}^{\frac{\mathrm{1}}{\mathrm{3}}} \:\mathrm{and}\:\mathrm{q}^{\frac{\mathrm{1}}{\mathrm{3}}} \mathrm{each}\:\mathrm{has}\:\mathrm{actually}\:\right. \\ $$$$\left.\mathrm{3}\:\mathrm{values}\right) \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{p}^{\frac{\mathrm{1}}{\mathrm{3}}}…
Question Number 30456 by daffa123 last updated on 22/Feb/18 $${proof}\:{that} \\ $$$$\frac{{a}^{\mathrm{2}} }{\left({a}−{b}\right)\left({a}−{c}\right)}+\frac{{b}^{\mathrm{2}} }{\left({b}−{c}\right)\left({b}−{a}\right)}+\frac{{c}^{\mathrm{2}} }{\left({c}−{a}\right)\left({c}−{b}\right)}=\:{a}+{b}+{c} \\ $$ Answered by Rasheed.Sindhi last updated on 22/Feb/18 $$\frac{{a}^{\mathrm{2}}…
Question Number 161527 by mnjuly1970 last updated on 19/Dec/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 161507 by cortano last updated on 18/Dec/21 Answered by MathsFan last updated on 19/Dec/21 $${x}=\pm\mathrm{0}.\mathrm{51} \\ $$ Answered by mr W last updated…
Question Number 95964 by i jagooll last updated on 29/May/20 $$\begin{cases}{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{13}}\\{\mathrm{2x}^{\mathrm{2}} +\mathrm{3y}=\mathrm{2xy}^{\mathrm{2}} }\end{cases} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 95967 by pticantor last updated on 29/May/20 Commented by pticantor last updated on 29/May/20 $${please}\:{i}\:{need}\:{help}\:{please}\:{please} \\ $$ Answered by john santu last updated…
Question Number 161500 by HongKing last updated on 18/Dec/21 $$\mathrm{x}^{\mathrm{6}} \:-\:\mathrm{6x}^{\mathrm{5}} \:+\:\mathrm{ax}^{\mathrm{4}} \:+\:\mathrm{bx}^{\mathrm{3}} \:+\:\mathrm{cx}^{\mathrm{2}} \:+\:\mathrm{dx}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{all}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{are}\:\mathrm{positive} \\ $$$$\mathrm{find}\:\:\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=? \\ $$ Answered by mr W…