Question Number 194846 by dimentri last updated on 17/Jul/23 $$\:\:\:\:\:\:\underbrace{\:} \\ $$ Answered by som(math1967) last updated on 17/Jul/23 $$\:{x}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{10}+\mathrm{6}\sqrt{\mathrm{3}}}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\:\sqrt{\mathrm{5}+\mathrm{1}+\mathrm{2}\sqrt{\mathrm{5}}}−\sqrt{\mathrm{5}}} \\ $$$${x}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{3}.\mathrm{3}+\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{1}}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\:\sqrt{\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)^{\mathrm{2}} }−\sqrt{\mathrm{5}}} \\ $$$${x}=\frac{\sqrt[{\mathrm{3}}]{\left(\sqrt{\mathrm{3}}\right)^{\mathrm{3}}…
Question Number 194853 by horsebrand11 last updated on 17/Jul/23 Answered by cortano12 last updated on 17/Jul/23 $$\:\:\:\underbrace{\Subset} \\ $$ Commented by horsebrand11 last updated on…
Question Number 194851 by Guriya last updated on 17/Jul/23 $$\boldsymbol{\mathrm{Name}}\:\:\:\boldsymbol{\mathrm{Zainab}}\:\boldsymbol{\mathrm{Bibi}} \\ $$$$\boldsymbol{\mathrm{BC}}\mathrm{200400692} \\ $$$$\boldsymbol{\mathrm{A}}\mathrm{ssignmeng}\:\mathrm{No}#\mathrm{2}\: \\ $$$$\boldsymbol{\mathrm{Mth}}\:\mathrm{621} \\ $$$$\mathrm{solution}.. \\ $$$$\mathrm{a}_{\mathrm{n}} =\frac{\left(\mathrm{n}− \right)!}{\left(\mathrm{n}+ \right)^{\mathrm{2}} } \\…
Question Number 194844 by sniper237 last updated on 16/Jul/23 $$\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} +\left({x}+\mathrm{3}\right)^{\mathrm{2}} =\left({x}^{\mathrm{2}} +{ax}+{b}\right)\left({x}^{\mathrm{2}} +{cx}+{d}\right) \\ $$$${Find}\:{a},{b},{c},{d}. \\ $$ Answered by Frix last updated on…
Question Number 194809 by dimentri last updated on 16/Jul/23 $$\:\:\:\:{If}\:{f}\left({x}\right)={ax}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{3}\:{and}\: \\ $$$$\:\:\:{g}\left({x}\right)=\mathrm{3}{x}−\mathrm{3}\:{intersection}\:{at} \\ $$$$\:{points}\:\left(\mathrm{1},{h}\right)\:{and}\:\left(\mathrm{3},{t}\right). \\ $$$$\:\:{Find}\:\:\underline{ } \\ $$ Answered by horsebrand11 last updated…
Question Number 194837 by mr W last updated on 16/Jul/23 $${for}\:{x}>\mathrm{0}\:{find}\:{the}\:{minimum}\:{of}\:{the} \\ $$$${function}\:{f}\left({x}\right)={x}^{\mathrm{3}} +\frac{\mathrm{5}}{{x}}. \\ $$ Answered by Frix last updated on 16/Jul/23 $${f}'\left({x}\right)=\mathrm{0} \\…
Question Number 194808 by York12 last updated on 15/Jul/23 $$ \\ $$$${suppose}\:{a},{b},{c}\:{are}\:{positive}\:{real}\:{numbers} \\ $$$${prove}\:{the}\:{inequality} \\ $$$$\left(\frac{{a}+{b}}{\mathrm{2}}\right)\left(\frac{{b}+{c}}{\mathrm{2}}\right)\left(\frac{{c}+{a}}{\mathrm{2}}\right)\geqslant\left(\frac{{a}+{b}+{c}}{\mathrm{3}}\right)\sqrt[{\mathrm{3}}]{\left({abc}\right)^{\mathrm{2}} } \\ $$ Commented by York12 last updated on…
Question Number 194779 by sniper237 last updated on 15/Jul/23 $${If}\:\:{a}\:\:{divided}\:{by}\:{b}\:{gives}\:{q}\:\:{remaining}\:{r} \\ $$$${Then}\:\:\frac{{a}}{{b}}\:=\:{q},{rrr}…\:\:{in}\:{base}\:{b}+\mathrm{1} \\ $$ Commented by Frix last updated on 15/Jul/23 No problem �� Commented by Frix…
Question Number 194756 by horsebrand11 last updated on 15/Jul/23 $$\:\:\:\:\:\:\underbrace{\boldsymbol{{b}}} \\ $$ Answered by cortano12 last updated on 15/Jul/23 $$\:\:\frac{\mathrm{x}−\mathrm{a}}{\:\sqrt{\mathrm{x}}+\sqrt{\mathrm{a}}}\:=\:\frac{\mathrm{x}−\mathrm{a}}{\mathrm{3}\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{a}}\right)}\:+\mathrm{2}\sqrt{\mathrm{a}} \\ $$$$\:\frac{\mathrm{3}\left(\mathrm{x}−\mathrm{a}\right)}{\mathrm{3}\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{a}}\right)}−\frac{\left(\mathrm{x}−\mathrm{a}\right)}{\mathrm{3}\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{a}}\right)}\:=\:\mathrm{2}\sqrt{\mathrm{a}} \\ $$$$\:\:\frac{\mathrm{2}\left(\mathrm{x}−\mathrm{a}\right)}{\mathrm{3}\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{a}}\right)}\:=\:\mathrm{2}\sqrt{\mathrm{a}} \\…
Question Number 194791 by dimentri last updated on 15/Jul/23 $$\:\:\:\underline{\underbrace{\boldsymbol{{x}}}} \\ $$ Answered by Frix last updated on 15/Jul/23 $$\mathrm{If}\:\sqrt[{\mathrm{7}}]{−{r}}=−\sqrt[{\mathrm{7}}]{{r}} \\ $$$$\frac{{x}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2}{x}^{\mathrm{2}} }\sqrt[{\mathrm{7}}]{{x}−\sqrt{\mathrm{2}}}=\frac{{x}^{\frac{\mathrm{9}}{\mathrm{7}}} }{\mathrm{2}\sqrt[{\mathrm{7}}]{{x}+\sqrt{\mathrm{2}}}}…