Question Number 228571 by malwan last updated on 23/Apr/26 $${x}\:=\:\:^{\mathrm{3}} \sqrt{\mathrm{16}}\:+\:\:^{\mathrm{3}} \sqrt{\mathrm{36}}\:−\:\:^{\mathrm{3}} \sqrt{\mathrm{24}} \\ $$$$\frac{\mathrm{10}}{{x}^{\mathrm{6}} }\:−\:\frac{{x}^{\mathrm{3}} }{\mathrm{10}^{\mathrm{3}} }\:−\:\frac{\mathrm{30}}{{x}^{\mathrm{3}} }\:=\:? \\ $$ Commented by Frix last…
Question Number 228520 by Prax last updated on 19/Apr/26 Commented by mr W last updated on 22/Apr/26 $${a}_{\mathrm{1}} =\mathrm{2} \\ $$$${a}_{\mathrm{2}} =\frac{\mathrm{1}+\mathrm{2}}{\mathrm{3}−\mathrm{2}}=\mathrm{3} \\ $$$${a}_{\mathrm{3}} =\frac{\mathrm{1}+\mathrm{3}}{\mathrm{3}−\mathrm{3}}=\frac{\mathrm{4}}{\mathrm{0}}\:!!!…
Question Number 228095 by gregori last updated on 19/Mar/26 $$\:\:\: {a}_{{n}+\mathrm{2}} \:=\:{a}_{{n}} \:+\:{a}_{{n}+\mathrm{1}} \: {n}\geqslant\mathrm{1}\:. \\ $$$$\:\: {a}_{\mathrm{7}} \:=\:\mathrm{120}\: {a}_{\mathrm{8}} \:=?\: \\ $$ Commented by…
Question Number 227898 by needothink last updated on 26/Feb/26 $$\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$$$=>\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{x}}}\:=\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}}\:+\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{b}}}\:+\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}} \\ $$$$=\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{x}}}\:−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\:=\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}}\:+\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{b}}} \\ $$$$=\:\frac{\boldsymbol{\mathrm{x}}\:−\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{x}}\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{x}}\right)}\:=\:\frac{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{ab}}} \\ $$$$=\:\frac{\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{ax}}+\boldsymbol{\mathrm{bx}}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\:=\:\frac{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{ab}}} \\ $$$$=\:\frac{−\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}\:}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{ax}}+\boldsymbol{\mathrm{bx}}\:}\:=\:\frac{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{ab}}} \\ $$$$=\:\frac{−\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\right)\:}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{ax}}+\boldsymbol{\mathrm{bx}}\:}\:=\:\frac{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{ab}}}…
Question Number 227824 by mr W last updated on 20/Feb/26 $$\mathrm{A}\:\mathrm{team}\:\mathrm{that}\:\mathrm{is}\:\mathrm{100}\:\mathrm{meters}\:\mathrm{long}\:\mathrm{is} \\ $$$$\mathrm{moving}\:\mathrm{forward}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\: \\ $$$$\mathrm{at}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{speed}.\:\mathrm{A}\:\mathrm{messenger}\: \\ $$$$\mathrm{runs}\:\mathrm{at}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{speed}\:\mathrm{from}\:\mathrm{the}\: \\ $$$$\mathrm{rear}\:\mathrm{of}\:\mathrm{the}\:\mathrm{team}\:\mathrm{to}\:\mathrm{the}\:\mathrm{front}\:\mathrm{to}\: \\ $$$$\mathrm{deliver}\:\mathrm{a}\:\mathrm{message}.\:\mathrm{Then}\:\mathrm{without}\: \\ $$$$\mathrm{changing}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{he}\:\mathrm{runs}\:\mathrm{back}\:\mathrm{to}\: \\ $$$$\mathrm{the}\:\mathrm{rear}\:\mathrm{of}\:\mathrm{the}\:\mathrm{team}.\:\mathrm{By}\:\mathrm{the}\:\mathrm{time}\:\mathrm{he}\:…
Question Number 227139 by Spillover last updated on 02/Jan/26 Answered by MrAjder last updated on 03/Jan/26 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}\left(\mathrm{2}{n}−\mathrm{1}\right)}\overset{\left[\mathrm{1}\right]} {=}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{n}}\right) \\ $$$$\overset{\left[\mathrm{1}\right]} {=}\underset{{n}=\mathrm{1}}…
Question Number 226943 by Spillover last updated on 20/Dec/25 Commented by fantastic2 last updated on 26/Dec/25 $${the}\:{rectangle}\:{is}\:{not}\:{unique} \\ $$ Commented by fantastic2 last updated on…
Question Number 226942 by Spillover last updated on 20/Dec/25 Answered by Spillover last updated on 24/Dec/25 Answered by Spillover last updated on 24/Dec/25 Answered by…
Question Number 226919 by Spillover last updated on 19/Dec/25 Answered by Ghisom_ last updated on 19/Dec/25 $$\underset{{n}} {\underbrace{\mathrm{6}…}}\mathrm{8}^{\mathrm{2}} =\underset{{n}} {\underbrace{\mathrm{4}…}}\mathrm{6}\underset{{n}} {\underbrace{\mathrm{2}…}}\mathrm{4} \\ $$$$\Rightarrow\:\underset{{digits}} {\sum}=\left(\mathrm{4}+\mathrm{2}\right){n}+\mathrm{6}+\mathrm{4}=\mathrm{6}{n}+\mathrm{10} \\…
Question Number 226798 by aba_math last updated on 15/Dec/25 $${let}\:{gcd}\left({m},{n}\right)=\mathrm{1}.\:{Determine}\:{gcd}\left(\mathrm{5}^{{m}} +\mathrm{7}^{{m}} ,\mathrm{5}^{{n}} +\mathrm{7}^{{n}} \right) \\ $$ Answered by Frix last updated on 15/Dec/25 $${a},\:{b},\:{c},\:…\:\in\mathbb{N} \\…