Question Number 127968 by liberty last updated on 03/Jan/21 $$\:\mathrm{How}\:\mathrm{many}\:\mathrm{x}\epsilon\mathbb{R}\:\mathrm{satisfy}\:\sqrt[{\mathrm{7}}]{\mathrm{x}}\:−\sqrt[{\mathrm{5}}]{\mathrm{x}}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{x}}\:−\sqrt{\mathrm{x}}\: \\ $$ Answered by MJS_new last updated on 03/Jan/21 $${x}=\mathrm{0}\vee{x}\approx.\mathrm{0117154773}\vee{x}=\mathrm{1} \\ $$ Terms of Service…
Question Number 62428 by Tawa1 last updated on 21/Jun/19 Commented by mathsolverby Abdo last updated on 21/Jun/19 $${this}\:{question}\:{is}\:{done}\:{take}\:{a}\:{look}\:{at}\:{the} \\ $$$${platform} \\ $$ Commented by Tawa1…
Question Number 62424 by aliesam last updated on 21/Jun/19 Commented by mathmax by abdo last updated on 21/Jun/19 $$\mathrm{1}\:{is}\:{not}\:{root}\:{for}\:{this}\:{equatio}\: \\ $$$$\left({e}\right)\:\Leftrightarrow\frac{\mathrm{1}−{x}^{\mathrm{5}} }{\mathrm{1}−{x}}\:=\mathrm{0}\:\Leftrightarrow\:{x}^{\mathrm{5}} \:=\mathrm{1}\:\:{let}\:{x}\:={re}^{{i}\theta} \:\:\:\:\:\left(\:\:{we}\:{take}\:{x}\:{from}\:{C}\right) \\…
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Question Number 127924 by ajfour last updated on 03/Jan/21 $$\:\:\:\:{x}^{\mathrm{3}} ={x}+{c} \\ $$$${Solve}\:{for}\:{x},\:{even}\:{when}\:{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}. \\ $$ Commented by ajfour last updated on 03/Jan/21 $$\left({without}\:{trigonometric}\:{functions}\right) \\ $$…
Question Number 62372 by Rasheed.Sindhi last updated on 20/Jun/19 $${Solve}\:{for}\:{x}\:,\:{y} \\ $$$$\mathrm{3}{x}>\mathrm{2}{y}\:\wedge\:\mathrm{2}{x}<\mathrm{3}{y}\: \\ $$$${where}\:{x},{y}\in\mathbb{N} \\ $$ Answered by MJS last updated on 20/Jun/19 $$\frac{\mathrm{2}}{\mathrm{3}}{y}<{x}<\frac{\mathrm{3}}{\mathrm{2}}{y}\:\wedge\:{y}>\mathrm{0} \\…
Question Number 127908 by bramlexs22 last updated on 03/Jan/21 $$\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{1}.\mathrm{3}.\mathrm{5}}+\frac{\mathrm{3}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}}+\frac{\mathrm{4}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}.\mathrm{9}}+…=? \\ $$ Answered by liberty last updated on 03/Jan/21 $$\:\mathrm{Let}\:\lambda\:=\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{n}}{\mathrm{1}.\mathrm{3}.\mathrm{5}…\left(\mathrm{2n}+\mathrm{1}\right)}\: \\ $$$$\mathrm{consider}\:\frac{\mathrm{n}}{\mathrm{1}.\mathrm{3}.\mathrm{5}….\left(\mathrm{2n}+\mathrm{1}\right)}\:=\:\frac{\mathrm{n}}{\mathrm{1}.\left(\mathrm{3}.\mathrm{5}…\left(\mathrm{2n}−\mathrm{1}\right)\right).\left(\mathrm{2n}+\mathrm{1}\right)} \\…
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Question Number 62334 by smartsmith459@gmail.com last updated on 19/Jun/19 $${if}\:\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\:\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta\:{evaluate}\left(\alpha−\beta\right) \\ $$ Answered by Kunal12588 last updated on 20/Jun/19 $$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\left(\alpha+\beta\right)^{\mathrm{2}}…