Question Number 83935 by Roland Mbunwe last updated on 08/Mar/20 $${If}\:\:\boldsymbol{{x}}^{\mathrm{4}} \:{and}\:{higher}\:{powers}\:{of}\:{x}\:{are}\:{neglected},\:{show}\:{that} \\ $$$$\sqrt{\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}+{x}^{\mathrm{2}} }\right)=\mathrm{1}−{x}+\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{3}} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 18397 by geovane10math last updated on 20/Jul/17 $${Prove}\::\:\forall{n}\:\in\:\mathbb{N},\:{n}\:\geqslant\:\mathrm{2},\:{so}\:\exists\:{x},{y},{z}\:\mid\: \\ $$$${x},{y},{z}\:\in\:\mathbb{N}\:{such}\:{that}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{4}}{{n}}\:=\:\frac{\mathrm{1}}{{x}}\:+\:\frac{\mathrm{1}}{{y}}\:+\:\frac{\mathrm{1}}{{z}} \\ $$$${Example}: \\ $$$${choose}\:{n}\:=\:\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{4}}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{{x}}\:+\:\frac{\mathrm{1}}{{y}}\:+\:\frac{\mathrm{1}}{{z}} \\ $$$${If}\:{x}\:=\:\mathrm{1}\:\:,\:\:\:{y}\:=\:\mathrm{2}\:\:{and}\:\:\:{z}\:=\:\mathrm{2},\:{the}\:{equa}- \\ $$$${tion}\:{is}\:{correct}! \\…
Question Number 18379 by virus last updated on 19/Jul/17 Answered by mrW1 last updated on 19/Jul/17 $$\mathrm{T}=\mathrm{tension}\:\mathrm{in}\:\mathrm{rope} \\ $$$$\mathrm{4T}=\mathrm{m}_{\mathrm{A}} \mathrm{a}_{\mathrm{A}} \:\:\:…\left(\mathrm{i}\right) \\ $$$$\mathrm{F}−\mathrm{5T}=\mathrm{m}_{\mathrm{B}} \mathrm{a}_{\mathrm{B}} \:\:\:…\left(\mathrm{ii}\right)…
Question Number 18364 by 314159 last updated on 19/Jul/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 18363 by 314159 last updated on 19/Jul/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 18361 by diofanto last updated on 19/Jul/17 $${N}\:\mathrm{propositions}\:\mathrm{are}\:\mathrm{judged}\:\mathrm{by}\:\mathrm{2}{k}−\mathrm{1}\:\mathrm{people}. \\ $$$$\mathrm{Each}\:\mathrm{person}\:\mathrm{assigns}\:“\mathrm{true}''\:\mathrm{to} \\ $$$$\mathrm{exactly}\:{M}\:\mathrm{propositions}\:\mathrm{and}\:“\mathrm{false}'' \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{other}\:{N}−{M}\:\left({M}\:\leqslant\:{N}\right). \\ $$$$\mathrm{To}\:\mathrm{say}\:\mathrm{a}\:\mathrm{proposition}\:\mathrm{is}\:“\mathrm{approved}''\:\mathrm{means} \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{true}\:\mathrm{according}\:\mathrm{to}\:\mathrm{at}\:\mathrm{least}\:{k}\:\mathrm{judges}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{and}\:\mathrm{maximum}\:\mathrm{numbers} \\ $$$$\mathrm{of}\:\mathrm{approved}\:\mathrm{propositions}\:\mathrm{given}\:{N},\:{M}\:\mathrm{and}\:{k}. \\…
Question Number 18362 by 314159 last updated on 19/Jul/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 149425 by help last updated on 05/Aug/21 Answered by dumitrel last updated on 05/Aug/21 $${x}>\mathrm{0};{x}\neq\mathrm{1} \\ $$$$\sqrt[{\mathrm{4}}]{{log}_{\mathrm{3}} {x}}={y}>\mathrm{0}\Rightarrow{x}=\mathrm{3}^{{y}^{\mathrm{4}} } \\ $$$$\mathrm{9}^{{y}} −\mathrm{4}{x}^{\frac{\mathrm{1}}{{y}^{\mathrm{3}} }}…
Question Number 83886 by mr W last updated on 07/Mar/20 $${To}\:{the}\:{developers}\:{of}\:{TinkuTara}: \\ $$$${problem}\:\mathrm{1}: \\ $$$${i}\:{get}\:{no}\:{notifications}\:{when}\:{my}\:{posts} \\ $$$${are}\:{updated}. \\ $$$$ \\ $$$${problem}\:\mathrm{2}: \\ $$$${i}\:{can}\:{edit}\:{my}\:{post},\:{see}\:{picture}\:\mathrm{1},\:{but} \\ $$$${the}\:{content}\:{is}\:{not}\:{visiable},\:{see}\:{picture}\:\mathrm{2}.…
Question Number 18342 by mondodotto@gmail.com last updated on 19/Jul/17 Answered by alex041103 last updated on 19/Jul/17 $$\frac{{x}+\mathrm{4}}{{x}+\mathrm{1}}=\frac{{x}+\mathrm{1}}{{x}+\mathrm{1}}+\frac{\mathrm{3}}{{x}+\mathrm{1}}=\mathrm{1}+\frac{\mathrm{3}}{{x}+\mathrm{1}} \\ $$$$\frac{{x}−\mathrm{2}}{{x}−\mathrm{4}}=\frac{{x}−\mathrm{4}}{{x}−\mathrm{4}}+\frac{\mathrm{2}}{{x}−\mathrm{4}}=\mathrm{1}+\frac{\mathrm{2}}{{x}−\mathrm{4}} \\ $$$$\Rightarrow\frac{\mathrm{3}}{{x}+\mathrm{1}}<\frac{\mathrm{2}}{{x}−\mathrm{4}} \\ $$$${The}\:{inequality}\:{makes}\:{sense}\:{when} \\ $$$${x}\neq−\mathrm{1};\mathrm{4}…