Question Number 201200 by Calculusboy last updated on 01/Dec/23 Commented by Rasheed.Sindhi last updated on 02/Dec/23 $$\left(\:\left({x}−\mathrm{1}\right)^{{x}−\mathrm{1}} \:\right)^{\mathrm{1}/\mathrm{3}} =\left({x}−\mathrm{1}\right)^{\left({x}−\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} } \\ $$ Commented by Rasheed.Sindhi…
Question Number 201162 by Calculusboy last updated on 01/Dec/23 Answered by Frix last updated on 01/Dec/23 $${x}={y}={z}=\mathrm{1} \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 201190 by Calculusboy last updated on 01/Dec/23 Answered by mr W last updated on 01/Dec/23 $$\frac{\mathrm{1}}{\left(−{x}\right)^{\left(−{x}\right)} }=\frac{\mathrm{1}}{\mathrm{1000}} \\ $$$$\left(−{x}\right)^{\left(−{x}\right)} =\mathrm{1000} \\ $$$$\Rightarrow{x}=−\frac{\mathrm{4ln}\:\mathrm{10}}{{W}\left(\mathrm{4ln}\:\mathrm{10}\right)}\approx−\mathrm{5}.\mathrm{438583} \\…
Question Number 201152 by mathlove last updated on 01/Dec/23 $${if}\:\:\mathrm{4}^{{x}} +\mathrm{4}^{−{x}} =\mathrm{7} \\ $$$${then}\:\:\:\mathrm{8}^{{x}} +\mathrm{8}^{−{x}} =? \\ $$ Answered by BaliramKumar last updated on 01/Dec/23…
Question Number 201149 by Mingma last updated on 30/Nov/23 Answered by witcher3 last updated on 03/Dec/23 $$\mathrm{x}=\mathrm{n}\in\mathbb{N}\:\mathrm{y}=\frac{\mathrm{1}}{\mathrm{n}},\mathrm{z}=\frac{\mathrm{1}}{\mathrm{n}},\mathrm{n}\geqslant\mathrm{2} \\ $$$$\forall\mathrm{n}\in\mathbb{N}−\left\{\mathrm{0},\mathrm{1}\right\}\:\:\left(\mathrm{n},\frac{\mathrm{1}}{\mathrm{n}},\frac{\mathrm{1}}{\mathrm{n}}\right)\mathrm{is}\:\mathrm{solution} \\ $$$$ \\ $$ Terms of…
Question Number 201081 by Calculusboy last updated on 29/Nov/23 Answered by Rasheed.Sindhi last updated on 29/Nov/23 $$\mathrm{AnOther}\:\mathrm{Way} \\ $$$$\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} −\mathrm{22}{x}−\mathrm{24}=\mathrm{0} \\ $$$${let}\:{the}\:{two}\:{roots}\:{are}\:\mathrm{3}{a}\:\&\:\mathrm{4}{a}\:{where}\:{a}\neq\mathrm{0} \\ $$$$\bullet\mathrm{2}\left(\mathrm{3}{a}\right)^{\mathrm{3}}…
Question Number 201106 by mnjuly1970 last updated on 29/Nov/23 $$ \\ $$$$\:\:\:\:\:{calculate}\:… \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\zeta\left(\mathrm{2}{n}\:\right)}{\mathrm{2}^{\:{n}} .{n}}\:=\:? \\ $$$$ \\ $$ Answered by…
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Question Number 201041 by hardmath last updated on 28/Nov/23 $$\mathrm{4}\left(\mathrm{33}\right)\mathrm{7} \\ $$$$\mathrm{4}\left(\mathrm{24}\right)\mathrm{6} \\ $$$$\mathrm{5}\left(\:?\:\right)\mathrm{4} \\ $$$$ \\ $$$$\left.{a}\left.\right)\left.\mathrm{9}\left.\:\:\:\:\:{b}\right)\mathrm{18}\:\:\:\:\:{c}\right)\mathrm{27}\:\:\:\:\:{d}\right)\mathrm{36} \\ $$ Answered by Frix last updated…
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