Question Number 192172 by mehdee42 last updated on 10/May/23 $${Q}\mathrm{1}\:\therefore\:\:{x}=<\mathrm{1}{a}_{\mathrm{1}} {a}_{\mathrm{2}} …{a}_{{n}} >\in\mathbb{N}\:\:\&\:\:{y}=<{a}_{\mathrm{1}} {a}_{\mathrm{2}} …{a}_{{n}} \mathrm{1}>\in\mathbb{N} \\ $$$${if}\:\:{y}=\mathrm{3}{x}\:\:{then}\:\:,\:{find}\:{the}\:{smallest}\: \\ $$$${value}\:{of}\:\:{x} \\ $$$${Q}\mathrm{2}\:\therefore\:{with}\:{the}\:{above}\:{conditions}\:,{what}\:{other}\:{values}\: \\ $$$${can}\:{be}\:{placed}\:\:{besides}\:{the}\:{number}\:“\:\mathrm{1}\:''\: \\…
Question Number 192171 by mathlove last updated on 10/May/23 $$\mathrm{2}^{{x}^{{x}^{{x}} } } =\mathrm{2}^{\sqrt{\mathrm{2}}} \\ $$$${x}=? \\ $$ Commented by Frix last updated on 11/May/23 $$\mathrm{2}^{{x}^{{x}^{{x}}…
Question Number 126619 by O Predador last updated on 22/Dec/20 $$\: \\ $$$$\:\:\mathrm{5}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{7}^{\boldsymbol{\mathrm{x}}} \:=\:\left(\mathrm{9}.\mathrm{8}\right)^{\boldsymbol{\mathrm{x}}} \\ $$$$\: \\ $$$$\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$$$\: \\ $$ Answered by…
Question Number 192149 by Shrinava last updated on 09/May/23 $$\mathrm{Find}: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{3}} }\:+\:\frac{\mathrm{5}}{\mathrm{2}^{\mathrm{5}} }\:+\:\frac{\mathrm{7}}{\mathrm{2}^{\mathrm{7}} }\:+\:… \\ $$ Answered by aleks041103 last updated on 09/May/23 $$\underset{{k}=\mathrm{1}}…
Question Number 192143 by Shrinava last updated on 09/May/23 $$\mathrm{Find}: \\ $$$$\frac{\mathrm{7}}{\mathrm{2}}\:+\:\frac{\mathrm{77}}{\mathrm{22}}\:+\:\frac{\mathrm{777}}{\mathrm{222}}\:+\:\frac{\mathrm{7777}}{\mathrm{2222}}\:+…+\:\frac{\mathrm{77777777}}{\mathrm{22222222}} \\ $$ Commented by AST last updated on 09/May/23 $$=\frac{\mathrm{7}}{\mathrm{2}}×\mathrm{8}=\mathrm{28} \\ $$ Commented…
Question Number 192142 by mehdee42 last updated on 09/May/23 $${Question} \\ $$$${let}\:\:\:{x}=<{a}_{{n}} {a}_{{n}−\mathrm{1}} …{a}_{\mathrm{1}} {a}_{\mathrm{0}} >\:\in\mathbb{N}\:;\:{a}_{\mathrm{0}} \neq\mathrm{0}\:\:\&\: \\ $$$$\:{y}=<{a}_{{n}} {a}_{{n}−\mathrm{1}} …{a}_{\mathrm{1}} >\:\in\mathbb{N}\:\:{be}\: \\ $$$${two}\:{natural}\:{numbers}\: \\…
Question Number 192134 by mustafazaheen last updated on 09/May/23 $$\mathrm{when}\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{7}} =\sqrt{\mathrm{57125}}+\sqrt{\mathrm{57124}} \\ $$$$\mathrm{then}\:\:\:\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{7}} =? \\ $$ Answered by som(math1967) last updated on 09/May/23 $$\:\:\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)=\mathrm{1} \\…
Question Number 192087 by Mastermind last updated on 07/May/23 $$\mathrm{Let}\:\left\{\mathrm{H}_{\alpha} \right\}\:\in\:\Omega,\:\mathrm{be}\:\mathrm{a}\:\mathrm{family}\:\mathrm{of}\: \\ $$$$\mathrm{subgroup}\:\mathrm{of}\:\mathrm{a}\:\mathrm{group}\:\mathrm{G},\:\mathrm{then}\: \\ $$$$\mathrm{prove}\:\mathrm{that}\:\cap_{\alpha\:\in\:\Omega} \mathrm{H}_{\alpha} . \\ $$$$ \\ $$$$ \\ $$ Commented by…
Question Number 192062 by mehdee42 last updated on 07/May/23 $${prove}\:{it}\::\: \\ $$$$\:\:\:{times\_n}\:\:\:;\:\:\:\sqrt{\mathrm{4}+\sqrt{\mathrm{4}+\sqrt{\mathrm{4}+…+\sqrt{\mathrm{4}}}}\:\:}\:<\:\mathrm{3} \\ $$ Commented by ajfour last updated on 08/May/23 $${but}\:\mathrm{4}>\:\mathrm{3}\:\:\:\: \\ $$$${how}\:{can}\:{this}\:{be}… \\…
Question Number 192057 by Shrinava last updated on 07/May/23 $$\mathrm{Simplify}: \\ $$$$\frac{\sqrt{\mathrm{a}}\:\:+\:\:\sqrt{\mathrm{a}^{\mathrm{2}} }\:\:+\:\:\sqrt{\mathrm{a}^{\mathrm{3}} }\:\:+\:\:\sqrt{\mathrm{a}^{\mathrm{4}} }}{\left(\sqrt{\mathrm{a}}\:\:+\:\:\mathrm{1}\right)\centerdot\left(\mathrm{a}\:\:+\:\:\mathrm{1}\right)} \\ $$ Answered by mehdee42 last updated on 07/May/23 $$\:\:\:\frac{\sqrt{{a}}+{a}+{a}\sqrt{{a}}+{a}^{\mathrm{2}}…