Question Number 221544 by gregori last updated on 07/Jun/25 $$\:\:\sqrt[{{x}+\mathrm{1}}]{{x}−\mathrm{1}}\:=\:\sqrt[{{x}−\mathrm{1}}]{{x}+\mathrm{1}}\:,\:{x}\:{real}\: \\ $$ Answered by mr W last updated on 07/Jun/25 $${at}\:{first}\:{let}'{s}\:{have}\:{a}\:{look}\:{at}\:{the} \\ $$$${function}\:{f}\left({x}\right)={x}\mathrm{ln}\:{x}. \\ $$$${f}'\left({x}\right)=\mathrm{ln}\:{x}+\mathrm{1}=\mathrm{0}\:\Rightarrow{x}=\frac{\mathrm{1}}{{e}}…
Question Number 221504 by ajfour last updated on 07/Jun/25 $${Find}\:{x}\:{if}\:\:\:\:\frac{{x}^{\mathrm{4}} }{\mathrm{4}^{{x}} }=\frac{\mathrm{4}^{{x}} }{{x}^{\mathrm{4}} }\:\:.\:\:\:{x}\in\mathbb{R} \\ $$ Answered by mr W last updated on 07/Jun/25 $$\frac{{x}^{\mathrm{4}}…
Question Number 221430 by pronation last updated on 05/Jun/25 Answered by wewji12 last updated on 05/Jun/25 $$\mathrm{Automorphism}\:\mathrm{group}..?? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 221399 by Nicholas666 last updated on 02/Jun/25 $$ \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{let}}\:\boldsymbol{{a}},\boldsymbol{{b}}\:\geqslant\:\mathrm{0}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{{a}}\:+\:\boldsymbol{{b}}\:+\:\boldsymbol{{ab}}\:=\:\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}; \\ $$$$\:\:\frac{\mathrm{38}}{\mathrm{55}}\:\leqslant\:\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{2}} \:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{b}}^{\mathrm{2}} \:+\mathrm{2}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{2}} \:+\:\boldsymbol{{b}}^{\mathrm{2}} \:+\:\mathrm{1}}\:\leqslant\:\mathrm{1}\:\:,\:\:\:\:\:\:\: \\ $$$$\:\:\frac{\mathrm{463}}{\mathrm{812}}\:\leqslant\:\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{3}} \:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{b}}^{\mathrm{3}} \:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{3}} \:+\:\boldsymbol{{b}}^{\mathrm{3}}…
Question Number 221393 by universe last updated on 02/Jun/25 $$\:\:\:\:\:{a},\:{b},\:{c}\:{are}\:{complex}\:{number}\:{and}\: \\ $$$$\:\:\:\mid{a}\mid\:=\:\mid{b}\mid=\mid{c}\mid=\:\mathrm{1}\:{and}\:\:\frac{{a}^{\mathrm{2}} }{{bc}}+\frac{{b}^{\mathrm{2}} }{{ac}}\:+\frac{{c}^{\mathrm{2}} }{{ab}}\:=\:−\mathrm{1} \\ $$$$\:\:\:\:{where}\:\mid.\mid\:{is}\:\:{modules}\:{function} \\ $$$${then}\:\mid{a}+{b}+{c}\mid\:{can}\:{be}\: \\ $$$$\left({A}\right)\:\mathrm{0}\:\:\:\:\:\:\:\left({B}\right)\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\left({C}\right)\:\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\left({D}\right)\:\mathrm{2} \\ $$ Terms of…
Question Number 221377 by hardmath last updated on 01/Jun/25 $$\mathrm{Find}:\:\:\:\Omega\:=\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{2}\:\sqrt[{\boldsymbol{\mathrm{n}}}]{\mathrm{10}}\:−\:\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}} \:=\:? \\ $$ Answered by Ghisom last updated on 01/Jun/25 $$\mathrm{ln}\:\Omega\:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{n}\mathrm{ln}\:\left(\mathrm{2}×\mathrm{10}^{\mathrm{1}/{n}} −\mathrm{1}\right) \\…
Question Number 221359 by Engr_Jidda last updated on 31/May/25 Answered by Rasheed.Sindhi last updated on 31/May/25 $${x}\ast{y}=\mathrm{2}{x}+\mathrm{2}{y}−\frac{{xy}}{\mathrm{5}} \\ $$$${let}\:{e}\:{is}\:{an}\:{identity}\:{element}\:{of}\:\ast \\ $$$${x}\ast{e}={e}\ast{x}={x} \\ $$$$\Rightarrow\mathrm{2}{x}+\mathrm{2}{e}−\frac{{xe}}{\mathrm{5}}={x} \\ $$$${e}\left(\frac{\mathrm{10}−{x}}{\mathrm{5}}\right)=−{x}…
Question Number 221321 by dr1001sa last updated on 30/May/25 $${if}\:\mathrm{0}<{x}<{y}<{e}^{\mathrm{2}} \:{then} \\ $$$${y}^{\sqrt{{x}}} +{x}^{\mathrm{2}} +\mathrm{6}{xy}+\mathrm{18}{y}^{\mathrm{2}} +\frac{\mathrm{8}}{{x}}+\frac{\mathrm{16}}{\mathrm{9}{x}^{\mathrm{2}} {y}^{\mathrm{2}} }>\mathrm{2}+{x}^{\sqrt{{y}}} \\ $$ Terms of Service Privacy Policy…
Question Number 221322 by Frix last updated on 30/May/25 $$\mathrm{Solve}\:\mathrm{for}\:{x} \\ $$$${x}^{\frac{\mathrm{1}}{{a}}} +\sqrt{{x}^{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}} }={x}^{\frac{\mathrm{1}}{{b}}} \\ $$ Answered by fantastic last updated on 30/May/25 $${or}\:{x}^{\frac{\mathrm{1}}{{a}}} +\sqrt{{x}^{\frac{{a}+{b}}{{ab}}}…
Question Number 221245 by fantastic last updated on 28/May/25 $${Which}\:{is}\:{greater}\:\:\:\:\:\:\:\left(\sqrt{\mathrm{7}}+\mathrm{1}\right)\:\:{or}\:\:\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}}\right)????? \\ $$ Answered by mr W last updated on 28/May/25 $$\left(\sqrt{\mathrm{7}}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{8}+\mathrm{2}\sqrt{\mathrm{7}} \\ $$$$\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{8}+\mathrm{2}\sqrt{\mathrm{15}}>\mathrm{8}+\mathrm{2}\sqrt{\mathrm{7}}=\left(\sqrt{\mathrm{7}}+\mathrm{1}\right)^{\mathrm{2}}…