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} \\…
Question Number 226779 by Spillover last updated on 14/Dec/25 $${By}\:{using}\:{De}\:{Moivres}\:{theorm} \\ $$$${simplify} \\ $$$$\left({a}\right)\frac{\left(\mathrm{cos}\:\frac{\pi}{\mathrm{2}}−{i}\mathrm{sin}\:\frac{\pi}{\mathrm{2}}\right)\left(\mathrm{cos}\:\frac{\pi}{\mathrm{3}}+{i}\mathrm{sin}\:\frac{\pi}{\mathrm{3}}\right)}{\mathrm{cos}\:\frac{\pi}{\mathrm{3}}−{i}\mathrm{sin}\:\frac{\pi}{\mathrm{3}}} \\ $$$$\left({b}\right)\frac{\mathrm{cos}\:\frac{\pi}{\mathrm{8}}+{i}\mathrm{sin}\:\frac{\pi}{\mathrm{8}}}{\mathrm{cos}\:\frac{\pi}{\mathrm{6}}+{i}\mathrm{sin}\:\frac{\pi}{\mathrm{6}}} \\ $$ Answered by Frix last updated on 14/Dec/25…
Question Number 226775 by aba_math last updated on 14/Dec/25 $${Prove}\:{that}\:\left({a}−{b}\right)\left({a}−{c}\right)\left({a}−{d}\right)\left({b}−{c}\right)\left({b}−{d}\right)\left({c}−{d}\right)\:{divisible}\:{by}\:\mathrm{12} \\ $$ Terms of Service Privacy Policy Contact: [email protected]
Question Number 226721 by Spillover last updated on 11/Dec/25 Answered by breniam last updated on 12/Dec/25 $${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xyy}'=\mathrm{0} \\ $$$${t}=\mathrm{ln}\mid{x}\mid \\ $$$${x}={e}^{{t}} \\ $$$${y}\left({x}\right)={z}\left({t}\right)…