Question Number 79256 by ajfour last updated on 23/Jan/20 $$\mathrm{3}{s}^{\mathrm{2}} −\mathrm{2}{ps}−\mathrm{3}{cp}−\mathrm{1}=\mathrm{0}\:\:\:{and} \\ $$$$\mathrm{3}{s}−\mathrm{2}{p}−{sp}^{\mathrm{2}} −\mathrm{3}{cp}^{\mathrm{2}} =\mathrm{0} \\ $$$${find}\:{s}\:{and}\:{p}\:{both}\:{real}\:{in}\:{terms} \\ $$$${of}\:{c}\:\in\mathbb{R}. \\ $$ Answered by MJS last…
Question Number 79249 by Pratah last updated on 23/Jan/20 Commented by mr W last updated on 23/Jan/20 $${can}\:{you}\:{tell}\:{according}\:{to}\:{the}\:{definition} \\ $$$${in}\:{your}\:{question} \\ $$$$\left[\mathrm{0}.\mathrm{8}\right]=\mathrm{0} \\ $$$$\left[−\mathrm{0}.\mathrm{8}\right]=?\:\: \\…
Question Number 144768 by ajfour last updated on 04/Jul/21 $$\:\:{x}^{\mathrm{3}} −{x}−{c}=\mathrm{0} \\ $$$${let}\:\:{x}=\sqrt{{t}+{p}}+\sqrt{{t}+{q}} \\ $$$$\left({t}+{p}\right)\sqrt{{t}+{p}}+\left({t}+{q}\right)\sqrt{{t}+{q}} \\ $$$$+\mathrm{3}\left({t}+{p}\right)\sqrt{{t}+{q}}+\mathrm{3}\left({t}+{q}\right)\sqrt{{t}+{p}} \\ $$$$−\sqrt{{t}+{p}}−\sqrt{{t}+{q}}−{c}=\mathrm{0} \\ $$$${let}\:\:\left({t}+{q}\right)+\mathrm{3}\left({t}+{p}\right)−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{4}{t}=\mathrm{1}−\left({p}+{q}\right) \\ $$$$\Rightarrow\:\:\mathrm{4}{t}+\mathrm{4}{p}=\mathrm{3}{p}−{q}+\mathrm{1}…
Question Number 79233 by jagoll last updated on 23/Jan/20 $$\frac{\mathrm{1}}{\mathrm{x}\left(\mathrm{x}+\mathrm{1}\right)}+\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}+\mathrm{2}\right)}+ \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{x}+\mathrm{3}\right)}\leqslant\frac{\mathrm{3}}{\mathrm{4}} \\ $$ Commented by john santu last updated on 23/Jan/20 $$\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{x}+\mathrm{1}}+\frac{\mathrm{1}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{{x}+\mathrm{2}} \\ $$$$+\frac{\mathrm{1}}{{x}+\mathrm{2}}−\frac{\mathrm{1}}{{x}+\mathrm{3}}\leqslant\frac{\mathrm{3}}{\mathrm{4}}…
Question Number 144760 by mathdanisur last updated on 28/Jun/21 $$\sqrt[{\mathrm{3}}]{\mathrm{1}+\sqrt{{x}}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{1}-\sqrt{{x}}}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{5}} \\ $$$${Find}\:\:\boldsymbol{{x}}=? \\ $$ Answered by Olaf_Thorendsen last updated on 28/Jun/21 $$\mathrm{Let}\:\alpha\:=\:\sqrt[{\mathrm{3}}]{\mathrm{1}+\sqrt{{x}}}\:\mathrm{and}\:\beta\:=\:\sqrt[{\mathrm{3}}]{\mathrm{1}−\sqrt{{x}}} \\ $$$$\alpha+\beta\:\:=\:\sqrt[{\mathrm{3}}]{\mathrm{5}} \\…
Question Number 144767 by mnjuly1970 last updated on 29/Jun/21 Answered by Rasheed.Sindhi last updated on 29/Jun/21 $$\left({x}^{\mathrm{3}} −\frac{\mathrm{2}}{{x}^{\mathrm{2}} }\right)^{{m}} ={a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} {x}^{\mathrm{2}} +…+{a}_{{n}} {x}^{{n}}…
Question Number 144742 by loveineq last updated on 28/Jun/21 $$\mathrm{Let}\:{a},{b},{c}>\mathrm{0}\:\mathrm{and}\:{a}+{b}+{c}\:=\:\mathrm{3}.\:\mathrm{Prove}\:\mathrm{that}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\frac{{ab}}{{ab}+\mathrm{1}}+\frac{{bc}}{{bc}+\mathrm{1}}+\frac{{ca}}{{ca}+\mathrm{1}}}{\frac{\mathrm{1}}{{ab}+\mathrm{1}}+\frac{\mathrm{1}}{{bc}+\mathrm{1}}+\frac{\mathrm{1}}{{ca}+\mathrm{1}}}\:\geqslant\:{abc} \\ $$$$\left(\mathrm{Found}\:\mathrm{by}\:\mathrm{WolframAlpha}\:\mathrm{and}\:\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{inspired}\:\mathrm{by}\:\mathrm{my}\:\mathrm{old}\:\mathrm{problem}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 144736 by mathdanisur last updated on 28/Jun/21 $${P}\left({z}\right)={az}^{\mathrm{3}} +{z}^{\mathrm{2}} −\left({a}+\mathrm{6}\right){z}+{b}−\mathrm{6} \\ $$$${P}\left({z}\right)=\left({z}^{\mathrm{2}} +\mathrm{4}\right)\centerdot{Q}\left({z}\right) \\ $$$${Find}\:\:{a}\centerdot{b}=? \\ $$ Answered by nimnim last updated on…
Question Number 79190 by mr W last updated on 23/Jan/20 $${if}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{50}, \\ $$$${find}\:{the}\:{minimum}\:{and}\:{maximum}\:{of} \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{8}\left({x}+{y}\right)+\mathrm{20} \\ $$ Commented by jagoll last updated…
Question Number 144716 by mathdanisur last updated on 28/Jun/21 $$\int\left(\frac{{dx}}{{e}^{\mathrm{2}{x}} +\mathrm{1}}\right)=? \\ $$ Answered by imjagoll last updated on 28/Jun/21 $$\:\mathrm{let}\:\mathrm{e}^{\mathrm{x}} \:=\:\mu\:\Rightarrow\mathrm{dx}\:=\:\frac{\mathrm{d}\mu}{\mathrm{e}^{\mathrm{x}} }\:=\:\frac{\mathrm{d}\mu}{\mu} \\ $$$$\mathrm{I}=\int\:\left(\frac{\mathrm{1}}{\mu^{\mathrm{2}}…