Question Number 183864 by Michaelfaraday last updated on 31/Dec/22 Commented by MJS_new last updated on 31/Dec/22 $$\left.{b}\right) \\ $$ Commented by Michaelfaraday last updated on…
Question Number 118322 by mr W last updated on 16/Oct/20 $${solve} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${with}\:{x},{y},{z}\in\mathbb{N} \\ $$ Commented by malwaan last updated on 17/Oct/20 $$\frac{{yz}+{xz}+{xy}}{{xyz}}\:=\:\frac{\mathrm{3}}{\mathrm{4}}\:\left({x}\neq\mathrm{0};{y}\neq\mathrm{0};{z}\neq\mathrm{0}\right)…
Question Number 183848 by Michaelfaraday last updated on 30/Dec/22 Commented by MJS_new last updated on 31/Dec/22 $${y}=−{a}\vee{y}=−{b} \\ $$$$\mathrm{look}\:\mathrm{at}\:\mathrm{it},\:\mathrm{it}'\mathrm{s}\:\mathrm{plain}\:\mathrm{to}\:\mathrm{see} \\ $$ Answered by a.lgnaoui last…
Question Number 183847 by Michaelfaraday last updated on 30/Dec/22 Answered by MJS_new last updated on 31/Dec/22 $$\mathrm{5}{x}^{\mathrm{2}} −\mathrm{7}{x}={t} \\ $$$$\sqrt{{t}+\mathrm{10}}−\sqrt{{t}−\mathrm{9}}=\mathrm{1} \\ $$$$\sqrt{\mathrm{100}}−\sqrt{\mathrm{81}}=\mathrm{1} \\ $$$${t}=\mathrm{90} \\…
Question Number 183846 by Michaelfaraday last updated on 30/Dec/22 Answered by MJS_new last updated on 31/Dec/22 $$\mathrm{1}.\:{x}=\mathrm{0}\:\Rightarrow\:\sqrt{{x}−\sqrt{{x}}}=\mathrm{0} \\ $$$$\mathrm{2}.\:{x}>\mathrm{0} \\ $$$${t}=\sqrt{{x}−\sqrt{{x}}}\geqslant\mathrm{0}\:\Rightarrow\:{x}=\frac{\left(\mathrm{1}\pm\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\mathrm{insert}\:\mathrm{in}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation}\:\mathrm{to}\:\mathrm{get}…
Question Number 183843 by Michaelfaraday last updated on 30/Dec/22 Answered by mr W last updated on 31/Dec/22 $$\left(\underset{{a}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}{C}_{{a}} ^{\mathrm{4}} {x}^{\mathrm{2}{a}} \right)\left(\underset{{b}=\mathrm{0}} {\overset{\mathrm{7}} {\sum}}{C}_{{b}}…
Question Number 118298 by hmh2k20 last updated on 16/Oct/20 $$\:\:\:\:\:\mathrm{b},\:\mathrm{x},\:\mathrm{y},\:\mathrm{c}\:\mathrm{are}\:\mathrm{consecutive}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{a}\:\mathrm{G}.\mathrm{P}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:\:\:\:\:\boldsymbol{\mathrm{x}}\:=\:\boldsymbol{\mathrm{br}},\:\boldsymbol{\mathrm{y}}\:=\:\boldsymbol{\mathrm{br}}^{\mathrm{2}} ,\:\boldsymbol{\mathrm{c}}\:=\:\boldsymbol{\mathrm{br}}^{\mathrm{3}} \: \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\mathrm{A}.\mathrm{M}.\:\mathrm{between}\:\mathrm{b}\:\mathrm{and}\:\mathrm{c}\:\mathrm{is}\:\mathrm{a} \\ $$$$\:\:\:\:\:\:\:\therefore\:\:\:\:\:\frac{\mathrm{b}+\mathrm{c}}{\mathrm{2}}\:\:=\:\mathrm{a}\:\:\:\:\Rightarrow\:\:\mathrm{2}\boldsymbol{\mathrm{a}}\:=\:\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}} \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\boldsymbol{\mathrm{a}}\mathrm{bc}\:=\:\left(\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\right)\mathrm{bc} \\…
Question Number 183835 by a.lgnaoui last updated on 30/Dec/22 $${Resoudre}\:{le}\:{systeme} \\ $$$$\:\:\mathrm{abc}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{30} \\ $$$$\:\:\:\mathrm{a}+\mathrm{b}+\mathrm{c}\:\:\:\:\:\:\:=\mathrm{10} \\ $$$$\:\:\:\mathrm{ab}+\mathrm{bc}+\mathrm{ac}=\mathrm{31} \\ $$$$ \\ $$ Answered by Frix last updated…
Question Number 183825 by Michaelfaraday last updated on 30/Dec/22 $${solve}\:{for}\:{x}\:{by}\:{using}\:{lambert}\:{function} \\ $$$$\mathrm{2}^{{x}} +\mathrm{3}{x}=\mathrm{8} \\ $$ Answered by MJS_new last updated on 30/Dec/22 $$\mathrm{generally} \\ $$$${b}^{{x}}…
Question Number 118286 by 1549442205PVT last updated on 16/Oct/20 $$\mathrm{What}\:\mathrm{condition}\:\mathrm{should}\:\mathrm{be}\:\mathrm{satisfied}\:\mathrm{by} \\ $$$$\mathrm{the}\:\mathrm{vectors}\:\:\boldsymbol{\mathrm{a}}\:\mathrm{and}\:\:\boldsymbol{\mathrm{b}}\:\mathrm{for}\:\mathrm{the}\:\mathrm{following}\: \\ $$$$\mathrm{relations}\:\mathrm{to}\:\mathrm{hold}\:\mathrm{true}\::\left(\mathrm{a}\right)\mid\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\mid=\mid\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}\mid \\ $$$$;\left(\mathrm{b}\right)\mid\:\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\mid>\mid\:\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}\mid;\left(\mathrm{c}\right)\mid\boldsymbol{\mathrm{a}}+\:\boldsymbol{\mathrm{b}}\mid<\:\mid\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}\mid \\ $$ Answered by TANMAY PANACEA last updated on…