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 $$\:\mathrm{let}\:\mathrm{e}^{\mathrm{x}} \:=\:\mu\:\Rightarrow\mathrm{dx}\:=\:\frac{\mathrm{d}\mu}{\mathrm{e}^{\mathrm{x}} }\:=\:\frac{\mathrm{d}\mu}{\mu} \\ $$$$\mathrm{I}=\int\:\left(\frac{\mathrm{1}}{\mu^{\mathrm{2}}…
Question Number 13647 by chux last updated on 22/May/17 $$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{5}…..\left(\mathrm{1}\right) \\ $$$$\mathrm{3x}^{\mathrm{2}} +\mathrm{xy}+\mathrm{y}^{\mathrm{2}} =\mathrm{1}…..\left(\mathrm{2}\right) \\ $$$$ \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{find}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y} \\ $$ Commented…
Question Number 13645 by chux last updated on 21/May/17 $$\mathrm{please}\:\mathrm{is}\:\mathrm{factor}\:\mathrm{theorem}\:\mathrm{and}\:\mathrm{error} \\ $$$$\mathrm{and}\:\mathrm{trial}\:\mathrm{the}\:\mathrm{same}?\:\mathrm{please}\:\mathrm{help}\: \\ $$$$\mathrm{cause}\:\mathrm{i}\:\mathrm{think}\:\mathrm{theres}\:\mathrm{a}\:\mathrm{difference} \\ $$$$\mathrm{but}\:\mathrm{i}\:\mathrm{cant}\:\mathrm{explain}\:\mathrm{it}. \\ $$$$ \\ $$$$\mathrm{Thankz}. \\ $$ Terms of Service…
Question Number 144708 by mathdanisur last updated on 28/Jun/21 $${Simplify}:\:\:\left(\frac{\mathrm{1}}{\left({x}-\mathrm{2}\right)!}\:-\:\frac{\mathrm{1}}{\left({x}-\mathrm{1}\right)!}\right)\centerdot{x}! \\ $$ Answered by MJS_new last updated on 28/Jun/21 $$\left(\frac{\left({x}−\mathrm{1}\right){x}}{{x}!}−\frac{{x}}{{x}!}\right){x}!={x}^{\mathrm{2}} −\mathrm{2}{x}={x}\left({x}−\mathrm{2}\right) \\ $$ Commented by…