Question Number 155377 by mathdanisur last updated on 29/Sep/21 $$\mathrm{For}\:\mathrm{two}\:\mathrm{non}-\mathrm{negative}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\mathrm{a}^{\mathrm{6}} \:+\:\mathrm{b}^{\mathrm{6}} \:=\:\mathrm{2} \\ $$$$\mathrm{Find}\:\:\mathrm{a};\mathrm{b}\:\:\mathrm{such}\:\mathrm{that}\:\:\mathrm{3}\left(\mathrm{a}+\mathrm{b}\right)=\mathrm{1}+\mathrm{5}\sqrt{\mathrm{ab}} \\ $$ Answered by MJS_new last updated on 30/Sep/21…
Question Number 155372 by mathdanisur last updated on 29/Sep/21 Commented by Rasheed.Sindhi last updated on 30/Sep/21 $${What}'{s}\:{meant}\:{by}\:\mathrm{Z}.\mathrm{Q}\:? \\ $$ Commented by mathdanisur last updated on…
Question Number 89834 by Cheyboy last updated on 19/Apr/20 $${Find}\:{x}\: \\ $$$$\boldsymbol{{e}}^{\boldsymbol{{x}}} =\:\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{1} \\ $$$$\boldsymbol{{anyother}}\:\boldsymbol{{method}}\:\boldsymbol{{apart}}\:\boldsymbol{{from}}\:\boldsymbol{{N}}{ewton}'{s} \\ $$ Commented by Joel578 last updated on 19/Apr/20…
Question Number 89829 by student work last updated on 19/Apr/20 Commented by MJS last updated on 19/Apr/20 English please / Français s'il vous plaît Commented by student work last updated on…
Question Number 89826 by I want to learn more last updated on 19/Apr/20 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\:\:\:\:\:\:\mathrm{x}^{\mathrm{2}} \:\:+\:\:\mathrm{xy}\:\:+\:\:\mathrm{y}^{\mathrm{2}} \:\:\:=\:\:\:\mathrm{7}\:\:\:\:\:\:\:\:\:……\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\mathrm{y}^{\mathrm{2}} \:\:+\:\:\mathrm{yz}\:\:+\:\:\mathrm{z}^{\mathrm{2}} \:\:\:=\:\:\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:……\:\left(\mathrm{ii}\right) \\ $$$$\:\:\:\:\:\:\mathrm{z}^{\mathrm{2}} \:\:+\:\:\mathrm{xz}\:\:+\:\:\mathrm{x}^{\mathrm{2}}…
Question Number 155362 by mathdanisur last updated on 29/Sep/21 $$\mathrm{if}\:\:\mathrm{a};\mathrm{b}\geqslant\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{a}^{\mathrm{4}} \:+\:\mathrm{b}^{\mathrm{4}} \:=\:\mathrm{17}\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\mathrm{15}\left(\mathrm{a}\:+\:\mathrm{b}\right)\:\geqslant\:\mathrm{17}\:+\:\mathrm{14}\sqrt{\mathrm{2ab}} \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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} \\…