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Category: Algebra

someone-help-cheking-this-x-3-x-c-let-x-p-2q-q-2p-p-q-p-3-8q-3-6pq-p-2q-q-3-8p-3-6pq-q-2p-3-p-2q-q-2p-p-2q-q-2p-p-2q-q-2p-c-7-p-3-q-3-6pq

Question Number 128090 by ajfour last updated on 04/Jan/21 $${someone}\:{help}\:{cheking}\:{this}: \\ $$$$\:\:\:\:{x}^{\mathrm{3}} ={x}+{c}\:\: \\ $$$${let}\:\:{x}=\left({p}−\mathrm{2}{q}\right)+\left({q}−\mathrm{2}{p}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:−\left({p}+{q}\right) \\ $$$${p}^{\mathrm{3}} −\mathrm{8}{q}^{\mathrm{3}} −\mathrm{6}{pq}\left({p}−\mathrm{2}{q}\right)+ \\ $$$${q}^{\mathrm{3}} −\mathrm{8}{p}^{\mathrm{3}} −\mathrm{6}{pq}\left({q}−\mathrm{2}{p}\right)+…

Question-128065

Question Number 128065 by 0731619177 last updated on 04/Jan/21 Commented by mnjuly1970 last updated on 04/Jan/21 $${lim}_{{n}\rightarrow\infty\:\:\:} \underset{{k}=\mathrm{1}\:} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$${lim}_{{n}\rightarrow\infty} \left(\underset{{k}=\mathrm{1}}…

n-0-1-n-8n-3-

Question Number 128048 by bramlexs22 last updated on 04/Jan/21 $$\:\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{8n}+\mathrm{3}}\:=? \\ $$ Answered by Olaf last updated on 04/Jan/21 $$\mathrm{L}\left(\lambda,\alpha,{s}\right)\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{e}^{\mathrm{2}{i}\pi\lambda{n}}…

solve-for-x-2-x-2-x-2-x-2-x-3-

Question Number 62489 by Tawa1 last updated on 21/Jun/19 $$\mathrm{solve}\:\mathrm{for}\:\mathrm{x}:\:\:\:\:\:\:\:\frac{\sqrt{\mathrm{2}\:−\:\mathrm{x}}\:+\:\sqrt{\mathrm{2}\:+\:\mathrm{x}}}{\:\sqrt{\mathrm{2}\:−\:\mathrm{x}}\:−\:\sqrt{\mathrm{2}\:+\:\mathrm{x}}}\:\:=\:\:\mathrm{3} \\ $$ Answered by som(math1967) last updated on 22/Jun/19 $$\frac{\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{2}+{x}}+\sqrt{\mathrm{2}−{x}}−\sqrt{\mathrm{2}+{x}}}{\:\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{2}+{x}}−\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{2}+{x}}}=\frac{\mathrm{3}+\mathrm{1}}{\mathrm{3}−\mathrm{1}}\:\:\bigstar \\ $$$$\frac{\mathrm{2}\sqrt{\mathrm{2}−{x}}}{\mathrm{2}\sqrt{\mathrm{2}+{x}}}=\frac{\mathrm{4}}{\mathrm{2}} \\ $$$$\frac{\mathrm{2}−{x}}{\mathrm{2}+{x}}=\mathrm{4}\:\:\bigstar\bigstar \\…

Can-anyone-find-any-error-in-my-attempt-to-improve-upon-the-Cardano-s-cubic-formula-or-as-an-alternate-to-the-trigonometric-solution-here-it-goes-X-3-pX-q-0-let-X-p-x-and-with

Question Number 128007 by ajfour last updated on 03/Jan/21 $${Can}\:{anyone}\:{find}\:{any}\:{error}\:{in} \\ $$$${my}\:{attempt}\:{to}\:{improve}\:{upon} \\ $$$${the}\:{Cardano}'{s}\:{cubic}\:{formula}.. \\ $$$${or}\:{as}\:{an}\:{alternate}\:{to}\:{the} \\ $$$${trigonometric}\:{solution}… \\ $$$${here}\:{it}\:{goes}: \\ $$$$\:\:\:\:{X}^{\mathrm{3}} −{pX}−{q}=\mathrm{0} \\ $$$${let}\:\:\frac{{X}}{\:\sqrt{{p}}}\:=\:{x}\:;\:\:{and}\:{with}\:\frac{{q}}{{p}\sqrt{{p}}}\:=\:{c}\:,…

Question-62468

Question Number 62468 by bshahid010@gmail.com last updated on 21/Jun/19 Commented by mathmax by abdo last updated on 21/Jun/19 $${let}\:\mathrm{3}^{{x}} \:={t}\:\:\:\Rightarrow\left({k}−\mathrm{2}\right){t}^{\mathrm{2}} −\mathrm{2}{t}\:+\mathrm{3}>\mathrm{0}\:\:\:\:\forall{t}>\mathrm{0}\:\Rightarrow\:\Delta^{'} <\mathrm{0}\:\Rightarrow \\ $$$$\left(−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3}\left({k}−\mathrm{2}\right)<\mathrm{0}\:\Rightarrow\mathrm{1}−\mathrm{3}{k}\:+\mathrm{6}\:<\mathrm{0}\:\Rightarrow\mathrm{7}−\mathrm{3}{k}\:<\mathrm{0}\:\Rightarrow{k}>\frac{\mathrm{7}}{\mathrm{3}}…

Find-the-number-of-digit-in-2-50-

Question Number 62449 by Tawa1 last updated on 21/Jun/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{digit}\:\mathrm{in}\:\:\:\:\mathrm{2}^{\mathrm{50}} \\ $$ Commented by Tawa1 last updated on 21/Jun/19 $$\mathrm{And}\:\mathrm{can}\:\mathrm{we}\:\mathrm{find}\:\mathrm{a}\:\mathrm{general}\:\mathrm{nth}\:\mathrm{number}\:\mathrm{of}\:\mathrm{term}\:\mathrm{in}\:\mathrm{any}\:\mathrm{number}\:\mathrm{and}\:\:\mathrm{powers} \\ $$ Answered by Rasheed.Sindhi…

Find-the-remainder-when-2014-is-divisible-by-2017-

Question Number 62452 by Tawa1 last updated on 21/Jun/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when}\:\:\:\mathrm{2014}!\:\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\:\mathrm{2017} \\ $$ Answered by Rasheed.Sindhi last updated on 21/Jun/19 $${Wilson}'{s}\:{Theorm}: \\ $$$$\:\:\:\:\:\:\:\left({p}−\mathrm{1}\right)!\equiv−\mathrm{1}\left({mod}\:{p}\right)\::\:{p}\in\mathbb{P} \\ $$$$\:\:\:\because\:\:\:\:\mathrm{2017}\in\mathbb{P} \\…