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Category: Algebra

2-5-6-2-

Question Number 58936 by hhghg last updated on 01/May/19 $$\mathrm{2}+\left\{\left[\mathrm{5}+\mathrm{6}\right]\right\}×\mathrm{2} \\ $$ Answered by Tony Lin last updated on 01/May/19 $$=\mathrm{2}+\mathrm{11}×\mathrm{2}=\mathrm{24} \\ $$ Terms of…

6h-18-

Question Number 58932 by hhghg last updated on 01/May/19 $$\mathrm{6h}=\mathrm{18} \\ $$ Commented by Tawa1 last updated on 01/May/19 $$\mathrm{Divide}\:\mathrm{both}\:\mathrm{side}\:\mathrm{by}\:\:\mathrm{6} \\ $$$$\:\:\:\:\frac{\mathrm{6h}}{\mathrm{6}}\:\:=\:\:\frac{\mathrm{18}}{\mathrm{6}} \\ $$$$\:\:\:\:\mathrm{h}\:\:=\:\:\mathrm{3} \\…

reposting-this-x-8-8x-7-16x-6-208x-5-152x-4-928x-3-704x-2-1088x-368-0-nobody-wants-to-try-it-s-beautiful-

Question Number 58915 by MJS last updated on 01/May/19 $$\mathrm{reposting}\:\mathrm{this}: \\ $$$${x}^{\mathrm{8}} −\mathrm{8}{x}^{\mathrm{7}} −\mathrm{16}{x}^{\mathrm{6}} +\mathrm{208}{x}^{\mathrm{5}} −\mathrm{152}{x}^{\mathrm{4}} −\mathrm{928}{x}^{\mathrm{3}} +\mathrm{704}{x}^{\mathrm{2}} +\mathrm{1088}{x}−\mathrm{368}=\mathrm{0} \\ $$$$\mathrm{nobody}\:\mathrm{wants}\:\mathrm{to}\:\mathrm{try}?\:\mathrm{it}'\mathrm{s}\:\mathrm{beautiful}… \\ $$ Commented by…

Solve-for-x-x-x-x-729-

Question Number 58899 by Tawa1 last updated on 01/May/19 $$\boldsymbol{\mathrm{S}}\mathrm{olve}\:\mathrm{for}\:\:\mathrm{x}:\:\:\:\:\:\mathrm{x}^{\mathrm{x}^{\mathrm{x}} \:\:} =\:\:\mathrm{729} \\ $$ Commented by mr W last updated on 01/May/19 $${you}\:{can}'{t}\:{solve}\:{this}\:{using}\:{Lambert}\:{function}. \\ $$$${but}\:{you}\:{can}\:{solve}\:{x}^{{x}^{{x}^{{x}}…

Question-189975

Question Number 189975 by mathlove last updated on 25/Mar/23 Answered by a.lgnaoui last updated on 27/Mar/23 $${Q}_{\mathrm{1}} \:\:\:\frac{{lnx}}{\mathrm{2}−{x}}=−\frac{{lnx}}{{x}\left(\mathrm{1}−\frac{\mathrm{2}}{{x}}\right)}=\frac{{lnx}}{{x}}×\left(\frac{{x}}{\mathrm{2}−{x}}\right) \\ $$$${U}^{'} =\frac{{lnx}}{{x}}\:\:\:\:\:\:\:\:\:\:\rightarrow{U}=\frac{\mathrm{1}}{\mathrm{2}}\left({lnx}\right)^{\mathrm{2}} \\ $$$${V}=\frac{{x}}{\mathrm{2}−{x}}\:\:\:\:\:\:\:\:\:\:\rightarrow{V}'=\frac{\mathrm{2}}{\left(\mathrm{2}−{x}\right)^{\mathrm{2}} } \\…