Question Number 24286 by Joel577 last updated on 15/Nov/17 $$\mathrm{If}\:\mid{x}\mid\:<\:\mathrm{1}\:\mathrm{then} \\ $$$$\left({x}\:+\:\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{4}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{8}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{16}} \:+\:\mathrm{1}\right)….. \\ $$$$\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$ Answered by mrW1 last updated…
Question Number 155359 by mathlove last updated on 29/Sep/21 Commented by mathlove last updated on 30/Sep/21 $${pleas}\:{answer} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 155344 by mathdanisur last updated on 29/Sep/21 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{integers}: \\ $$$$ \\ $$$$\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{3x}\left(\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{y}\:-\:\mathrm{1}\right)\:+\:\mathrm{4y}^{\mathrm{2}} \:+\:\mathrm{4y}\:-\:\mathrm{6}\:=\:\mathrm{0} \\ $$ Answered by ghimisi last updated on…
Question Number 155345 by mathdanisur last updated on 29/Sep/21 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{in}\:\mathbb{R} \\ $$$$\frac{\mathrm{5}\sqrt{\mathrm{x}+\mathrm{1}}}{\:\sqrt{\mathrm{1}\:-\:\mathrm{x}\:+\:\mathrm{x}^{\mathrm{2}} }\:+\:\mathrm{2}\sqrt{\mathrm{x}\:+\:\mathrm{1}}}\:=\:\mathrm{4x}^{\mathrm{2}} \:-\:\mathrm{5x}\:+\:\mathrm{5} \\ $$ Commented by MJS_new last updated on 29/Sep/21 $$\mathrm{no}\:\mathrm{real}\:\mathrm{solution} \\…
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Question Number 155335 by mathdanisur last updated on 28/Sep/21 $$\mathrm{2x}^{\mathrm{5}} \:+\:\mathrm{3x}^{\mathrm{4}} \:-\:\mathrm{7x}^{\mathrm{3}} \:-\:\mathrm{7x}^{\mathrm{2}} \:+\:\mathrm{3x}\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$$$\mathrm{x}_{\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}.\mathrm{5}} \:=\:? \\ $$ Answered by MJS_new last updated on…
Question Number 89794 by M±th+et£s last updated on 19/Apr/20 $${true}\:{or}\:{false} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{3}} }\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }\right)…..\left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\right)<\mathrm{3} \\ $$ Commented by mr W last updated on…
Question Number 89795 by A8;15: last updated on 19/Apr/20 Answered by MJS last updated on 19/Apr/20 $$\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}+\mathrm{1}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} =\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{1}=\left({x}^{\mathrm{2}}…
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Question Number 155295 by mathdanisur last updated on 28/Sep/21 $$\mathrm{Evaluate}:\:\:\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\frac{\Sigma\:\boldsymbol{\mathrm{n}}}{\boldsymbol{\mathrm{n}}^{\mathrm{2}} }\:=\:? \\ $$ Commented by mathdanisur last updated on 28/Sep/21 $$\mathrm{Very}\:\mathrm{nice}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{thank}\:\mathrm{you} \\ $$ Commented…