Question Number 128146 by sahnaz last updated on 04/Jan/21 $$\mathrm{4}<\mathrm{x}<\mathrm{5}\:\mathrm{a}=\sqrt{\mathrm{x}−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{x}−\mathrm{4}}\:\:}\:\:\:\mathrm{b}=\sqrt{\mathrm{x}−\mathrm{4}}−\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{x}−\mathrm{a}×\mathrm{b}=? \\ $$ Answered by Fikret last updated on 04/Jan/21 $${a}=\sqrt{{x}−\mathrm{4}}+\mathrm{1} \\ $$$${b}=\sqrt{{x}−\mathrm{4}}−\mathrm{1} \\ $$$${a}.{b}={x}−\mathrm{5} \\…
Question Number 128132 by I want to learn more last updated on 04/Jan/21 $$\mathrm{If}\:\:\:\mathrm{f}\left(\mathrm{x}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{x}}\right)\:\:\:=\:\:\:\mathrm{x}^{\mathrm{4}} \:\:+\:\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }\:,\:\:\:\:\:\mathrm{find}\:\:\mathrm{f}\left(\mathrm{5}\right) \\ $$ Answered by Olaf last updated on 04/Jan/21…
Question Number 62577 by Sr@2004 last updated on 23/Jun/19 Commented by Sr@2004 last updated on 23/Jun/19 $${please}\:{solve}\:\mathrm{12} \\ $$ Answered by som(math1967) last updated on…
Question Number 128090 by ajfour last updated on 04/Jan/21 $${someone}\:{help}\:{cheking}\:{this}: \\ $$$$\:\:\:\:{x}^{\mathrm{3}} ={x}+{c}\:\: \\ $$$${let}\:\:{x}=\left({p}−\mathrm{2}{q}\right)+\left({q}−\mathrm{2}{p}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:−\left({p}+{q}\right) \\ $$$${p}^{\mathrm{3}} −\mathrm{8}{q}^{\mathrm{3}} −\mathrm{6}{pq}\left({p}−\mathrm{2}{q}\right)+ \\ $$$${q}^{\mathrm{3}} −\mathrm{8}{p}^{\mathrm{3}} −\mathrm{6}{pq}\left({q}−\mathrm{2}{p}\right)+…
Question Number 128065 by 0731619177 last updated on 04/Jan/21 Commented by mnjuly1970 last updated on 04/Jan/21 $${lim}_{{n}\rightarrow\infty\:\:\:} \underset{{k}=\mathrm{1}\:} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$${lim}_{{n}\rightarrow\infty} \left(\underset{{k}=\mathrm{1}}…
Question Number 128048 by bramlexs22 last updated on 04/Jan/21 $$\:\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}}…
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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 $$\frac{\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{2}+{x}}+\sqrt{\mathrm{2}−{x}}−\sqrt{\mathrm{2}+{x}}}{\:\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{2}+{x}}−\sqrt{\mathrm{2}−{x}}+\sqrt{\mathrm{2}+{x}}}=\frac{\mathrm{3}+\mathrm{1}}{\mathrm{3}−\mathrm{1}}\:\:\bigstar \\ $$$$\frac{\mathrm{2}\sqrt{\mathrm{2}−{x}}}{\mathrm{2}\sqrt{\mathrm{2}+{x}}}=\frac{\mathrm{4}}{\mathrm{2}} \\ $$$$\frac{\mathrm{2}−{x}}{\mathrm{2}+{x}}=\mathrm{4}\:\:\bigstar\bigstar \\…
Question Number 128007 by ajfour last updated on 03/Jan/21 $${Can}\:{anyone}\:{find}\:{any}\:{error}\:{in} \\ $$$${my}\:{attempt}\:{to}\:{improve}\:{upon} \\ $$$${the}\:{Cardano}'{s}\:{cubic}\:{formula}.. \\ $$$${or}\:{as}\:{an}\:{alternate}\:{to}\:{the} \\ $$$${trigonometric}\:{solution}… \\ $$$${here}\:{it}\:{goes}: \\ $$$$\:\:\:\:{X}^{\mathrm{3}} −{pX}−{q}=\mathrm{0} \\ $$$${let}\:\:\frac{{X}}{\:\sqrt{{p}}}\:=\:{x}\:;\:\:{and}\:{with}\:\frac{{q}}{{p}\sqrt{{p}}}\:=\:{c}\:,…
Question Number 62468 by bshahid010@gmail.com last updated on 21/Jun/19 Commented by mathmax by abdo last updated on 21/Jun/19 $${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}}…