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…
Question Number 201047 by Mingma last updated on 28/Nov/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
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…
Question Number 201034 by Mingma last updated on 28/Nov/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 200971 by Mingma last updated on 27/Nov/23 Answered by AST last updated on 27/Nov/23 $${WLOG},{let}\:{a}\:{be}\:{the}\:{max}\:{element} \\ $$$${abcd}={a}+{b}+{c}+{d}\leqslant\mathrm{4}{a}\Rightarrow{bcd}\leqslant\mathrm{4} \\ $$$${bcd}=\mathrm{1}\Rightarrow{b}={c}={d}=\mathrm{1}\Rightarrow{a}+\mathrm{3}={a}\left({absurd}\right) \\ $$$${bcd}=\mathrm{2}\Rightarrow{b}+{c}+{d}=\mathrm{4}\Rightarrow\mathrm{4}+{a}=\mathrm{2}{a}\Rightarrow{a}=\mathrm{4} \\ $$$$\Rightarrow\left({a},{b},{c},{d}\right)=\left(\mathrm{4},\mathrm{2},\mathrm{1},\mathrm{1}\right);\left(\mathrm{4},\mathrm{1},\mathrm{2},\mathrm{1}\right);\left(\mathrm{4},\mathrm{1},\mathrm{1},\mathrm{2}\right)…
Question Number 200961 by jabarsing last updated on 27/Nov/23 $$\: \\ $$$$\:\:\:\:\:−{x}^{\mathrm{3}} +\mathrm{1}=\sqrt[{\mathrm{3}}]{−{x}+\mathrm{1}}\:\Rightarrow{x}=? \\ $$$$ \\ $$ Answered by cortano12 last updated on 27/Nov/23 Answered…
Question Number 200885 by cortano12 last updated on 26/Nov/23 $$\:\:\:\:\cancel{\underline{\underbrace{\boldsymbol{{x}}}}} \\ $$ Commented by cortano12 last updated on 26/Nov/23 $$\:\:\:\boldsymbol{\mathfrak{f}} \\ $$ Answered by Frix…