Question Number 798 by 123456 last updated on 15/Mar/15 $$\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+\centerdot\centerdot\centerdot}}}} \\ $$ Answered by prakash jain last updated on 15/Mar/15 $$\mathrm{Ramanujan}'{s}\:\mathrm{Formula} \\ $$$${x}+\mathrm{1}=\sqrt{\mathrm{1}+{x}\sqrt{\mathrm{1}+\left({x}+\mathrm{1}\right)\sqrt{\mathrm{1}+\left({x}+\mathrm{2}\right)\sqrt{\mathrm{1}+…}}}} \\ $$$$\mathrm{For}\:{x}=\mathrm{2}…
Question Number 66302 by ajfour last updated on 12/Aug/19 $${solved}\:{the}\:{general}\:{quintic}, \\ $$$${despite}\:{whatever}\:{proof}\:{that}\:{it} \\ $$$${cant}\:{be}\:{solved}\:{in}\:{a}\:{simple}\:{way}! \\ $$ Commented by mr W last updated on 12/Aug/19 $${i}\:{knew}\:{you}'{ll}\:{tell}\:{us}\:{the}\:{good}\:{news}.…
Question Number 66290 by hmamarques1994@gmail.com last updated on 12/Aug/19 $$\:\sqrt[{\sqrt{\mathrm{3}}}]{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\:+\:\boldsymbol{\mathrm{x}}^{\sqrt{\mathrm{3}}} \:−\:\mathrm{392}\:=\:\mathrm{0} \\ $$ Answered by Tanmay chaudhury last updated on 12/Aug/19 $$\left({x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} +{x}^{\sqrt{\mathrm{3}}}…
Question Number 749 by 123456 last updated on 06/Mar/15 $${x}=\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}\centerdot\centerdot\centerdot}}}}}} \\ $$$${y}=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\centerdot\centerdot\centerdot}}}}}} \\ $$$${wich}\:{statment}\:{is}\:{true}? \\ $$$${a}.\:{xy}=\mathrm{4}\vee{xy}=\mathrm{0}\vee{x}+{y}=\mathrm{4}\vee{x}+{y}=\mathrm{2} \\ $$$${b}.\:{x}\notin\mathbb{Z} \\ $$$${c}.{xy}\notin\mathbb{Z} \\ $$$${d}.{x}+{y}\notin\mathbb{Z} \\ $$ Answered…
Question Number 739 by malwaan last updated on 08/Mar/15 $${solve}\:{x}^{\mathrm{2}} −\mathrm{7}{y}^{\mathrm{2}} =\mathrm{1}\:{in}\:{Z} \\ $$ Commented by 123456 last updated on 06/Mar/15 $${S}=\left\{\left({x},{y}\right)\in\mathbb{Z}^{\mathrm{2}} \mid{x}^{\mathrm{2}} −\mathrm{7}{y}^{\mathrm{2}} =\mathrm{1}\right\}…
Question Number 733 by 123456 last updated on 05/Mar/15 $$\mid\mid\mid\mid\mid\mid\mid{x}^{\mathrm{2}} −{x}−\mathrm{1}\mid−\mathrm{3}\mid−\mathrm{5}\mid−\mathrm{7}\mid−\mathrm{9}\mid−\mathrm{11}\mid−\mathrm{13}\mid \\ $$$$={x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{48} \\ $$$${find}\:{all}\:{x}\:{real}\:{that}\:{is}\:{solution}\:{of}\:{above} \\ $$$${equation} \\ $$ Answered by prakash jain last…
Question Number 698 by 9999 last updated on 01/Mar/15 $${solve}\:{for}\:{x} \\ $$$$\frac{{a}}{{ax}−\mathrm{1}}+\frac{{b}}{{bx}−\mathrm{1}}={a}+{b} \\ $$ Commented by 123456 last updated on 28/Feb/15 $${a}=\mathrm{0} \\ $$$$\frac{{b}}{{bx}−\mathrm{1}}={b},{b}=\mathrm{0},\forall{x}\neq\frac{\mathrm{1}}{{b}} \\…
Question Number 669 by 112358 last updated on 21/Feb/15 $${Given}\:{only}\:{the}\:{standard}\:{result}\:\underset{{r}=\mathrm{1}} {\overset{{N}} {\sum}}{r}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{6}}{N}\left({N}+\mathrm{1}\right)\left(\mathrm{2}{N}+\mathrm{1}\right)\: \\ $$$${is}\:{applied}\:{to}\:{determining}\:{the}\:{series} \\ $$$$\mathrm{1}^{\mathrm{2}} +\mathrm{2}×\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{2}×\mathrm{4}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} +\mathrm{2}×\mathrm{6}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} +…+\mathrm{2}\left({n}−\mathrm{1}\right)^{\mathrm{2}} +{n}^{\mathrm{2}}…
Question Number 66199 by aliesam last updated on 10/Aug/19 $${calculate} \\ $$$${cos}\left(\mathrm{79}\right)=? \\ $$ Commented by $@ty@m123 last updated on 10/Aug/19 $${Use}\:{trigonometrical}\:{table}. \\ $$$${OR}\: \\…
Question Number 66197 by Rasheed.Sindhi last updated on 11/Aug/19 $$\mathrm{If}\:\:\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}=\mathrm{1},\mathrm{prove}\:\mathrm{that}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{n}} +\mathrm{x}^{\mathrm{n}−\mathrm{2}} +\mathrm{x}^{\mathrm{n}−\mathrm{4}} =\mathrm{0} \\ $$ Commented by mr W last updated on 10/Aug/19…