Question Number 195124 by Shlock last updated on 25/Jul/23 Answered by witcher3 last updated on 25/Jul/23 $$\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}\leqslant\sqrt{\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{y}+\mathrm{1}\right)} \\ $$$$\Leftrightarrow\mathrm{x}+\mathrm{y}+\mathrm{2}\sqrt{\mathrm{xy}}\leqslant\mathrm{xy}+\mathrm{x}+\mathrm{y}+\mathrm{1}\Leftrightarrow\mathrm{xy}+\mathrm{1}\geqslant\mathrm{2}\sqrt{\mathrm{xy}},\mathrm{AM}−\mathrm{GM} \\ $$$$\Rightarrow\forall\left(\mathrm{x},\mathrm{y}\right)\in\mathbb{R}_{+} \sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}\leqslant\sqrt{\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{y}+\mathrm{1}\right)} \\ $$$$\Rightarrow\forall\left(\mathrm{a},\mathrm{b}\right)\in\left[\mathrm{1},\infty\left[^{\mathrm{2}} \right.\right.…
Question Number 195121 by mathlove last updated on 25/Jul/23 $${x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}^{\mathrm{8}} +\mathrm{2}{x}^{\mathrm{7}} −\mathrm{47}{x}=? \\ $$ Answered by qaz last updated on 25/Jul/23 $${x}^{\mathrm{8}}…
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Question Number 195175 by valdirmd last updated on 25/Jul/23 Answered by Rasheed.Sindhi last updated on 26/Jul/23 $${x}^{\mathrm{4}} +{x}^{\mathrm{2}} =\mathrm{11}/\mathrm{5}\:,\:\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{\mathrm{1}/\mathrm{3}} =? \\ $$$$\blacktriangleright{Let}\:{y}=\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{\mathrm{1}/\mathrm{3}} \\…
Question Number 195116 by SajaRashki last updated on 24/Jul/23 $${hi}.\:{please}\:{represntation}\:\sqrt[{\mathrm{3}}]{\mathrm{2}}\:{on}\:{number}'{s}\:{axis} \\ $$$${with}\:{ruler}\:{and}\:{compass}\:{and}\:{pen}. \\ $$$${thank}\:{you} \\ $$ Commented by Frix last updated on 25/Jul/23 $$\mathrm{Impossible}. \\…
Question Number 195101 by mathlove last updated on 24/Jul/23 $$\sqrt{{ln}\mathrm{2}}\:\:\overset{?} {>}{ln}\mathrm{2} \\ $$ Answered by Frix last updated on 24/Jul/23 $$\mathrm{e}^{{x}} =\mathrm{2} \\ $$$$\mathrm{e}>\mathrm{2}\:\Rightarrow\:{x}<\mathrm{1} \\…
Question Number 195107 by Rupesh123 last updated on 24/Jul/23 Answered by JDamian last updated on 24/Jul/23 $$\mathrm{5}+\mathrm{15}+\mathrm{25}+\:\centerdot\centerdot\centerdot\:+\mathrm{975}+\mathrm{985}+\mathrm{995}= \\ $$$$=\frac{\mathrm{100}}{\mathrm{2}}\left(\mathrm{5}+\mathrm{995}\right)=\mathrm{50000} \\ $$ Terms of Service Privacy…
Question Number 195002 by mathlove last updated on 22/Jul/23 $${x}^{\sqrt{{x}}} =\left(\sqrt{{x}}\right)^{{x}} \:\:\:\:\:{x}=? \\ $$ Answered by dimentri last updated on 22/Jul/23 $$\:\:\:\left({x}\right)^{\sqrt{{x}}} \:=\:\left({x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}{x}} \\ $$$$\:\:\left(\mathrm{1}\right)\:\sqrt{{x}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{x}\Rightarrow{x}−\mathrm{2}\sqrt{{x}}\:=\mathrm{0}…
Question Number 195027 by York12 last updated on 22/Jul/23 $${a}_{\mathrm{3}} {x}^{\mathrm{3}} −{x}^{\mathrm{2}} +{a}_{\mathrm{1}} {x}−\mathrm{7}=\mathrm{0}\:{is}\:{a}\:{cubic}\:{polynomial}\:{in}\:{x} \\ $$$${whose}\:{Roots}\:{are}\:\alpha\:,\:\beta\:,\:\gamma\:{positive}\:{real}\:{numbers} \\ $$$${satisfying} \\ $$$$\frac{\mathrm{225}\alpha^{\mathrm{2}} }{\alpha^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{144}\beta^{\mathrm{2}} }{\beta^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{100}\gamma^{\mathrm{2}} }{\gamma^{\mathrm{2}}…
Question Number 195017 by mathlove last updated on 22/Jul/23 $$\left({x}+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{1}\:\:\:\:\:\:\:\:\:\:{x}=? \\ $$ Answered by Rasheed.Sindhi last updated on 22/Jul/23 $$\left({x}+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:{x}=? \\ $$$${Let}\:\:\:\:\:{x}+\mathrm{1}={y} \\…