Question Number 112097 by bemath last updated on 06/Sep/20 $$\:\:\:\:\:\:\mid\mid{x}−\mathrm{1}\mid+\mathrm{1}\mid\:<\:{x} \\ $$ Commented by mr W last updated on 06/Sep/20 $$\mid{x}−\mathrm{1}\mid+\mathrm{1}<{x} \\ $$$$\mid{x}−\mathrm{1}\mid<{x}−\mathrm{1} \\ $$$$\Rightarrow{no}\:{solution},\:{since}\:\mid{a}\mid\geqslant{a}.…
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Question Number 112087 by bobhans last updated on 06/Sep/20 $$\:\:\:\mathrm{a}\sqrt{\mathrm{a}}\:−\mathrm{3}\:=\:\mathrm{10}\sqrt{\mathrm{a}}\:\rightarrow\:\sqrt{\mathrm{a}}\:+\:\sqrt{\mathrm{a}^{−\mathrm{1}} }\:=? \\ $$ Answered by bemath last updated on 06/Sep/20 Answered by 1549442205PVT last updated…
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Question Number 46527 by Tawa1 last updated on 28/Oct/18 $$\underset{\mathrm{n}=\:\mathrm{1}} {\overset{\infty} {\boldsymbol{\sum}}}\:\left(\frac{\mathrm{log}\:\mathrm{n}}{\mathrm{n}}\right)^{\mathrm{2}} \\ $$$$\mathrm{Does}\:\mathrm{the}\:\mathrm{series}\:\mathrm{converge}\:\mathrm{or}\:\mathrm{diverge},\:\:\mathrm{help}\:\mathrm{find}\:\mathrm{the}\:\mathrm{sum}\:… \\ $$ Commented by maxmathsup by imad last updated on 28/Oct/18…
Question Number 112058 by mohssinee last updated on 05/Sep/20 Answered by MJS_new last updated on 06/Sep/20 $$\Rightarrow\:{x}=−\frac{\mathrm{19}\pm\sqrt{\mathrm{4}{a}^{\mathrm{2}} −\mathrm{407}}}{\mathrm{2}} \\ $$$$\mathrm{4}{a}^{\mathrm{2}} −\mathrm{407}={b}^{\mathrm{2}} \:\mathrm{with}\:{b}\in\mathbb{Z} \\ $$$$\mathrm{4}{a}^{\mathrm{2}} ={b}^{\mathrm{2}}…
Question Number 112053 by MJS_new last updated on 05/Sep/20 $$\mathrm{solve}\:\mathrm{for}\:{x},\:{y},\:{z}\:\in\mathbb{C}: \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}{x}=\sqrt{\mathrm{13}{x}^{\mathrm{2}} −\mathrm{52}{x}+\mathrm{40}} \\ $$$$\mathrm{6}{y}^{\mathrm{2}} −\mathrm{14}{x}=\sqrt{{x}^{\mathrm{2}} −\mathrm{220}{x}+\mathrm{300}} \\ $$$${z}^{\mathrm{2}} −\mathrm{2}{z}=\sqrt{−\mathrm{12}{x}^{\mathrm{2}} +\mathrm{72}{x}−\mathrm{132}} \\ $$$$\left[\mathrm{exact}\:\mathrm{solutions}\:\mathrm{possible}\:\mathrm{in}\:\mathrm{all}\:\mathrm{cases}\right] \\…
Question Number 177579 by mr W last updated on 07/Oct/22 $${if}\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} +\frac{{x}+{y}}{\mathrm{4}}=\frac{\mathrm{15}}{\mathrm{2}},\:{find}\:{maximum} \\ $$$${value}\:{of}\:{x}+{y}. \\ $$ Answered by TheHoneyCat last updated on 07/Oct/22 $$\mathrm{if}\:{x}={y}+\epsilon…
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