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Question Number 58084 by Tawa1 last updated on 17/Apr/19 Commented by MJS last updated on 17/Apr/19 $$\mathrm{sorry}\:\mathrm{but}\:\mathrm{do}\:\mathrm{you}\:\mathrm{really}\:\mathrm{believe}\:\mathrm{someone} \\ $$$$\mathrm{on}\:\mathrm{this}\:\mathrm{forum}\:\mathrm{could}\:\mathrm{be}\:\mathrm{able}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{problems} \\ $$$$\mathrm{which}\:\mathrm{can}'\mathrm{t}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{by}\:\mathrm{the}\:\mathrm{greatest} \\ $$$$\mathrm{mathematicians}\:\mathrm{of}\:\mathrm{our}\:\mathrm{time}? \\ $$…
Question Number 58077 by hhghg last updated on 17/Apr/19 $$\frac{\mathrm{3}}{\mathrm{10}}×\mathrm{2} \\ $$ Answered by $@ty@m last updated on 17/Apr/19 $${Pl}.\:{see}\:{my}\:{comment}\:{on}\:\mathrm{57882}. \\ $$ Commented by mr…
Question Number 189137 by Shrinava last updated on 12/Mar/23 $$\mathrm{It}\:\mathrm{is}\:\mathrm{known}\:\mathrm{that}\:\:\mathrm{x}\:\:\mathrm{is}\:\mathrm{rational} \\ $$$$\mathrm{x}\:\sqrt{\mathrm{28}\:+\:\mathrm{3}\sqrt{\mathrm{28}\:+\:\mathrm{3}\sqrt{\mathrm{28}\:+\:\mathrm{3}\sqrt{?}}}} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{dufference}\:\mathrm{of}\:\mathrm{possible}\:\mathrm{vaules} \\ $$$$\mathrm{of}\:\:\boldsymbol{\mathrm{x}} \\ $$ Commented by mr W last updated on…
Question Number 189133 by Shrinava last updated on 12/Mar/23 $$\mathrm{Convert}\:\mathrm{hexadecimal}\:\mathrm{number} \\ $$$$\mathrm{4}\:\mathrm{A}\:\mathrm{F}_{\mathrm{16}} \:\:\mathrm{to}\:\mathrm{decimal} \\ $$ Answered by mr W last updated on 12/Mar/23 $${digit}\:{A}=\mathrm{10} \\…
Question Number 189132 by Shrinava last updated on 12/Mar/23 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\:\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations}: \\ $$$$\begin{cases}{\mathrm{13a}+\mathrm{2b}+\mathrm{c}+\mathrm{6d}+\mathrm{2e}=\mathrm{96}}\\{\mathrm{5a}+\mathrm{9b}+\mathrm{2c}+\mathrm{7d}+\mathrm{3e}=\mathrm{75}}\\{\mathrm{7a}+\mathrm{8b}+\mathrm{17c}+\mathrm{11d}+\mathrm{7e}=\mathrm{99}}\\{\mathrm{3a}+\mathrm{3b}+\mathrm{3c}+\mathrm{d}+\mathrm{8e}=\mathrm{55}}\\{\mathrm{a}+\mathrm{7b}+\mathrm{6c}+\mathrm{4d}+\mathrm{9e}=\mathrm{79}}\end{cases} \\ $$ Answered by mr W last updated on 12/Mar/23 $${say}\:{f}={a}+{b}+{c}+{d}+{e}…
Question Number 189135 by Shrinava last updated on 12/Mar/23 Answered by a.lgnaoui last updated on 12/Mar/23 $$\left(\mathrm{257}−\mathrm{160}\right)=\left(\mathrm{264}−\mathrm{167}\right)=\left(\mathrm{271}−\mathrm{174}\right)=……=\left(\mathrm{2112}−\mathrm{2015}\right) \\ $$$$=\mathrm{97}\:\:\Rightarrow{suite}\:{aritmetique}\: \\ $$$${de}\:{raison}:\mathrm{97}\:\:\:;{u}_{\mathrm{0}} \:=\mathrm{257} \\ $$$${u}_{{n}+\mathrm{1}} =\mathrm{257}\left(\mathrm{1}+\mathrm{97}{n}\:\:\right)\:\left({n}>=\mathrm{0}\right)…
Question Number 189131 by Shrinava last updated on 12/Mar/23 Answered by cortano12 last updated on 12/Mar/23 $$\sqrt{\mathrm{19}+\mathrm{2}\sqrt{\mathrm{4}×\mathrm{15}}}\:=\mathrm{2}+\sqrt{\mathrm{15}} \\ $$ Answered by BaliramKumar last updated on…
Question Number 123595 by Snail last updated on 26/Nov/20 $${For}\:{real}\:{numbers}\:\left({a}/{b}/{c}\right)\:{define}\:{s}_{{n}} ={a}^{{n}} +{b}^{{n}} +{c}^{{n}} \\ $$$${Suppose}\:{s}_{\mathrm{1}} =\mathrm{2}\:{s}_{\mathrm{2}} =\mathrm{6}\:{and}\:{s}_{\mathrm{3}\:} =\mathrm{14}.{Prove}\:{that}\: \\ $$$$\mid{s}_{{n}} ^{\mathrm{2}} −{s}_{{n}−\mathrm{1}} {s}_{{n}+\mathrm{1}} \mid=\mathrm{8}={holds}\:\forall\:{n}>\mathrm{1} \\…
Question Number 189124 by Rupesh123 last updated on 12/Mar/23 Answered by witcher3 last updated on 15/Mar/23 $$\mathrm{x}+\mathrm{y}+\mathrm{x}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{xyz}} \\ $$$$\Leftrightarrow\mathrm{xyz}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{xyz}}\:;\mathrm{let}\:\mathrm{a}=\mathrm{xyz} \\ $$$$\Leftrightarrow\mathrm{a}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{a}}\Rightarrow\mathrm{a}^{\frac{\mathrm{2}}{\mathrm{3}}} \geqslant\mathrm{3}\Rightarrow\mathrm{a}\geqslant\sqrt[{\mathrm{2}}]{\mathrm{27}}=\mathrm{3}\sqrt{\mathrm{3}} \\ $$$$\mathrm{a}+\frac{\mathrm{1}}{\mathrm{a}}=\mathrm{f}\left(\mathrm{a}\right)\geqslant\mathrm{2},\forall\mathrm{a}>\mathrm{0} \\…