Question Number 187874 by mustafazaheen last updated on 23/Feb/23 $${how}\:{is}\:{solution} \\ $$$$\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{13}} =\mathrm{x}\:\:\:\:\:\:\:\:\:\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{221}} =? \\ $$$$\left.\mathrm{1}\left.\right)\left.\mathrm{x}^{−\mathrm{16}} \left.\:\:\:\:\:\:\:\:\:\:\mathrm{2}\right)\mathrm{x}^{−\mathrm{17}} \:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\right)\mathrm{x}^{\mathrm{221}} \:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}\right)\mathrm{x}^{\mathrm{21}} \\ $$ Answered by som(math1967) last…
Question Number 56800 by mr W last updated on 24/Mar/19 $${x},{y},{z}\:{are}\:{positive}\:{integers}. \\ $$$${find}\:{all}\:{solutions}\:{of}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}={xyz}. \\ $$ Commented by MJS last updated on 24/Mar/19 $$\mathrm{with}\:{z}=\mathrm{3}\:\mathrm{we}\:\mathrm{get}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{Fibonacci}\:\mathrm{numbers}…
Question Number 187871 by mustafazaheen last updated on 23/Feb/23 $${how}\:{is}\:{solution} \\ $$$$\mathrm{sgn}\left(\mathrm{cos}\frac{\mathrm{21}}{\mathrm{10}}\right)=? \\ $$ Answered by mr W last updated on 23/Feb/23 $$\frac{\mathrm{21}}{\mathrm{10}}=\mathrm{2}.\mathrm{1}\:>\frac{\pi}{\mathrm{2}}\:{but}\:<\pi \\ $$$$\Rightarrow\mathrm{cos}\:\frac{\mathrm{21}}{\mathrm{10}}\:<\mathrm{0}…
Question Number 187856 by Michaelfaraday last updated on 23/Feb/23 $${find}\:{x} \\ $$$$\mathrm{2}^{\sqrt{{x}}} =\mathrm{8}{x} \\ $$ Answered by mr W last updated on 23/Feb/23 $${t}=\sqrt{{x}}\:>\mathrm{0} \\…
Question Number 187811 by Rupesh123 last updated on 22/Feb/23 Answered by Frix last updated on 22/Feb/23 $${x}^{\mathrm{6}} +\frac{{x}^{\mathrm{5}} }{\mathrm{2}}−\frac{\mathrm{5}{x}^{\mathrm{4}} }{\mathrm{4}}−\frac{\mathrm{5}{x}^{\mathrm{3}} }{\mathrm{8}}+\frac{\mathrm{5}{x}^{\mathrm{2}} }{\mathrm{16}}+\frac{\mathrm{5}{x}}{\mathrm{32}}+\frac{\mathrm{1}}{\mathrm{64}}=\mathrm{0} \\ $$$$\mathrm{Try}\:\mathrm{to}\:\mathrm{find}\:\mathrm{2}\:\mathrm{cubic}\:\mathrm{factors}.\:\mathrm{If}\:\mathrm{it}'\mathrm{s}\:\mathrm{possible} \\…
Question Number 187780 by mnjuly1970 last updated on 21/Feb/23 $$ \\ $$$$\:\:{solve} \\ $$$$\:\:\:\:\lfloor\:\:\frac{\mathrm{1}}{\mathrm{4}\:}\:+\:\mathrm{2}^{\:{x}} \:\rfloor\:+\:\lfloor\:\frac{\mathrm{1}}{\mathrm{2}}\:+\:\mathrm{2}^{\:{x}+\mathrm{1}} \:\rfloor=\mathrm{1} \\ $$$$ \\ $$ Answered by witcher3 last updated…
Question Number 56705 by gunawan last updated on 22/Mar/19 $$\mathrm{the}\:\:\mathrm{smallest}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{S}=\left\{\:^{\mathrm{3}} \sqrt{{n}}−^{\mathrm{3}} \sqrt{{m}}\:\mid\:{n},\:{m}\:\in\mathbb{N}\right\}\:\:\mathrm{is}… \\ $$ Answered by kaivan.ahmadi last updated on 25/Mar/19 $${S}\:{has}\:{no}\:{smallest}\:{value}\:{of}\:\:{the} \\…
Question Number 56696 by Tawa1 last updated on 21/Mar/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{perpendicular}\:\mathrm{distance}\:\mathrm{from}\:\:\left(\mathrm{1},\:\mathrm{7},\:\mathrm{1}\right)\:\:\mathrm{to}\:\:\mathrm{3x}\:−\:\mathrm{2y}\:+\:\mathrm{2z}\:\:=\:\:\mathrm{6} \\ $$ Commented by mr W last updated on 21/Mar/19 $${d}=\frac{\mid\mathrm{3}×\mathrm{1}−\mathrm{2}×\mathrm{7}+\mathrm{2}×\mathrm{1}−\mathrm{6}\mid}{\:\sqrt{\mathrm{3}^{\mathrm{2}} +\left(−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }}=\frac{\mathrm{15}}{\:\sqrt{\mathrm{17}}} \\…
Question Number 56697 by Tawa1 last updated on 21/Mar/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{shortest}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{the}\:\mathrm{lines} \\ $$$$\:\:\:\mathrm{L}\:\:=\:\:\left(\mathrm{1},\:\mathrm{4},\:\mathrm{2}\right)\:+\:\mathrm{N}\left(\mathrm{1},\:\mathrm{3},\:\mathrm{2}\right)\:\:\:\mathrm{and} \\ $$$$\:\:\:\mathrm{r}\:\:=\:\:\left(−\mathrm{1},\:\mathrm{1},\:−\mathrm{1}\right)\:+\:\lambda\left(\mathrm{1},\:\mathrm{2},\:−\mathrm{1}\right) \\ $$ Answered by mr W last updated on 21/Mar/19 $${many}\:{methods}\:{to}\:{solve}.…
Question Number 56685 by pieroo last updated on 21/Mar/19 $$\mathrm{show}\:\mathrm{that}\:\alpha^{\mathrm{4}} +\beta^{\mathrm{4}} \:=\:\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \right)^{\mathrm{2}} \:−\mathrm{2}\alpha^{\mathrm{2}} \beta^{\mathrm{2}} \\ $$ Commented by pieroo last updated on 21/Mar/19…