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Category: Algebra

x-4-2-2-x-4-x-4-

Question Number 163222 by cortano1 last updated on 05/Jan/22 $$\:\:\:\sqrt{\frac{\sqrt{{x}+\mathrm{4}}\:+\mathrm{2}}{\mathrm{2}−\sqrt{{x}+\mathrm{4}}}}\:\leqslant\:{x}−\mathrm{4}\: \\ $$ Answered by blackmamba last updated on 05/Jan/22 $$\:{no}\:{solution}\:{for}\:{x}\in\mathbb{R} \\ $$$$\left(\mathrm{1}\right)\:{x}−\mathrm{4}>\mathrm{0}\Rightarrow{x}>\mathrm{4} \\ $$$$\left(\mathrm{2}\right)\:\frac{\sqrt{{x}+\mathrm{4}}+\mathrm{2}}{\mathrm{2}−\sqrt{{x}+\mathrm{4}}}\:\geqslant\mathrm{0}\:\:;\:\sqrt{{x}+\mathrm{4}}\:=\:{t} \\…

f-I-0-I-R-f-twice-derivable-f-f-continuous-f-x-f-x-f-x-2-x-I-then-prove-that-2f-x-y-2-f-x-f-y-x-y-I-

Question Number 163209 by HongKing last updated on 04/Jan/22 $$\mathrm{f}\::\:\mathrm{I}\:\rightarrow\:\left(\mathrm{0}\:;\:\infty\right)\:\:;\:\:\mathrm{I}\:\subset\:\mathbb{R} \\ $$$$\mathrm{f}\:-\:\mathrm{twice}\:\mathrm{derivable}\:\:;\:\:\mathrm{f}\:^{'} \:;\:\mathrm{f}\:^{''} \:-\:\mathrm{continuous} \\ $$$$\mathrm{f}\:^{''} \left(\mathrm{x}\right)\:\mathrm{f}\left(\mathrm{x}\right)\:\geqslant\:\left(\mathrm{f}\:^{'} \left(\mathrm{x}\right)\right)^{\mathrm{2}} \:;\:\:\forall\:\mathrm{x}\:\in\:\mathrm{I} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\mathrm{2f}\:\left(\frac{\mathrm{x}\:+\:\mathrm{y}}{\mathrm{2}}\right)\:\leqslant\:\mathrm{f}\left(\mathrm{x}\right)\:+\:\mathrm{f}\left(\mathrm{y}\right)\:\:;\:\:\forall\:\mathrm{x};\mathrm{y}\:\in\:\mathrm{I} \\ $$…

If-y-1-x-2-x-3-and-x-1-where-is-small-show-that-y-3-5-Hence-find-the-increase-in-y-when-x-is-increased-from-1-to-1-02-

Question Number 32110 by pieroo last updated on 19/Mar/18 $$\mathrm{If}\:\mathrm{y}=\mathrm{1}+\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{3}} \:\mathrm{and}\:\mathrm{x}=\mathrm{1}+\alpha,\:\mathrm{where}\:\alpha\:\mathrm{is}\:\mathrm{small},\:\mathrm{show} \\ $$$$\mathrm{that}\:\mathrm{y}\approx\mathrm{3}+\mathrm{5}\alpha.\:\mathrm{Hence},\:\mathrm{find}\:\mathrm{the}\:\mathrm{increase}\:\mathrm{in}\:\mathrm{y}\:\mathrm{when} \\ $$$$\mathrm{x}\:\mathrm{is}\:\mathrm{increased}\:\mathrm{from}\:\mathrm{1}\:\mathrm{to}\:\mathrm{1}.\mathrm{02} \\ $$ Answered by mrW2 last updated on 19/Mar/18…

Given-p-q-R-1-1-p-1-q-1-show-that-a-b-R-ab-a-p-p-b-q-q-

Question Number 97637 by Ar Brandon last updated on 09/Jun/20 $$\mathrm{Given}\:\mathrm{p},\mathrm{q}\in\mathbb{R}_{+} ^{\ast} −\left\{−\mathrm{1}\right\}/\frac{\mathrm{1}}{\mathrm{p}}+\frac{\mathrm{1}}{\mathrm{q}}=\mathrm{1}\:\mathrm{show}\:\mathrm{that}; \\ $$$$\forall\mathrm{a},\mathrm{b}\:\in\mathbb{R}\:\mathrm{ab}\leqslant\frac{\mathrm{a}^{\mathrm{p}} }{\mathrm{p}}+\frac{\mathrm{b}^{\mathrm{q}} }{\mathrm{q}} \\ $$ Commented by arcana last updated on…

Question-32099

Question Number 32099 by jasno91 last updated on 19/Mar/18 Answered by Joel578 last updated on 19/Mar/18 $$\mathrm{10}\frac{\mathrm{5}}{\mathrm{6}}\:−\:\mathrm{7}\frac{\mathrm{4}}{\mathrm{6}}\:+\:\mathrm{8}\frac{\mathrm{2}}{\mathrm{6}}\:−\:\mathrm{5}\frac{\mathrm{3}}{\mathrm{6}} \\ $$$$=\:\left(\mathrm{10}\:−\:\mathrm{7}\:+\:\mathrm{8}\:−\:\mathrm{5}\right)\:+\:\left(\frac{\mathrm{5}}{\mathrm{6}}\:−\:\frac{\mathrm{4}}{\mathrm{6}}\:+\:\frac{\mathrm{2}}{\mathrm{6}}\:−\:\frac{\mathrm{3}}{\mathrm{6}}\right) \\ $$$$=\:\:\mathrm{6}\:+\:\frac{\mathrm{0}}{\mathrm{6}} \\ $$$$=\:\mathrm{6} \\ $$…