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Question Number 150053 by mathdanisur last updated on 09/Aug/21 $$\mathrm{let}\:\:\mathrm{x};\mathrm{y};\mathrm{z};\mathrm{t}>\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{x}+\mathrm{y}+\mathrm{z}+\mathrm{t}=\mathrm{4} \\ $$$$\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{4}}{\left(\mathrm{xyzt}\right)^{\mathrm{2}} }\:+\:\mathrm{3}\:\geqslant\:\sqrt{\mathrm{45}\:+\:\mathrm{x}^{\mathrm{4}} \:+\:\mathrm{y}^{\mathrm{4}} \:+\:\mathrm{z}^{\mathrm{4}} \:+\:\mathrm{t}^{\mathrm{4}} } \\ $$ Terms of Service Privacy…
Question Number 150040 by bobhans last updated on 09/Aug/21 $$\mathrm{Let}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{be}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{such} \\ $$$$\mathrm{that}\:\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{1}\:.\mathrm{Prove}\:\mathrm{that}\: \\ $$$$\:\frac{\mathrm{ab}}{\mathrm{1}−\mathrm{c}^{\mathrm{2}} }\:+\frac{\mathrm{bc}}{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{ca}}{\mathrm{1}−\mathrm{b}^{\mathrm{2}} }\:\leqslant\:\frac{\mathrm{3}}{\mathrm{8}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 150039 by bobhans last updated on 09/Aug/21 $$\mathrm{Prove}\:\mathrm{that}\:\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{a}}\geqslant\sqrt{\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{1}}{\mathrm{b}^{\mathrm{2}} +\mathrm{1}}}+\sqrt{\frac{\mathrm{b}^{\mathrm{2}} +\mathrm{1}}{\mathrm{c}^{\mathrm{2}} +\mathrm{1}}}+\sqrt{\frac{\mathrm{c}^{\mathrm{2}} +\mathrm{1}}{\mathrm{a}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\mathrm{for}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{are}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{number}\: \\ $$ Answered by EDWIN88 last updated…
Question Number 18961 by Tinkutara last updated on 02/Aug/17 $$\mathrm{Find}\:\mathrm{arg}\left({z}\right),\:{z}\:=\:{i}^{{i}^{{i}} } . \\ $$ Answered by sma3l2996 last updated on 02/Aug/17 $${z}={i}^{\left({e}^{{i}\pi/\mathrm{2}} \right)^{{i}} } ={i}^{{e}^{−\pi/\mathrm{2}}…
Question Number 150029 by mathdanisur last updated on 08/Aug/21 $$\mathrm{Calcular}: \\ $$$$\sqrt{\mathrm{8}\:+\:\mathrm{2}\sqrt{\mathrm{8}\:+\:\mathrm{2}\sqrt{\mathrm{8}\:+\:…}}}\:=\:? \\ $$ Answered by nimnim last updated on 08/Aug/21 $$\mathrm{Let}\:\sqrt{\mathrm{8}+\mathrm{2}\sqrt{\mathrm{8}+\mathrm{2}\sqrt{\mathrm{8}+……}}}=\mathrm{x} \\ $$$$\Rightarrow\sqrt{\mathrm{8}+\mathrm{2x}}=\mathrm{x} \\…
Question Number 150034 by mathdanisur last updated on 08/Aug/21 $$\mathrm{Find}\:\boldsymbol{\mathrm{a}}\:\mathrm{closed}\:\mathrm{form}:\:\:\boldsymbol{\mathrm{a}}\in\mathbb{R}\:\:\mathrm{and}\:\:\boldsymbol{\mathrm{a}}\neq\mathrm{0} \\ $$$$\Omega\left(\mathrm{a}\right)=\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\frac{\mathrm{x}^{\mathrm{4}} }{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{a}^{\mathrm{4}} \mathrm{x}^{\mathrm{4}} \right)}\:\mathrm{dx} \\ $$ Answered by mathmax by abdo…
Question Number 150009 by mathdanisur last updated on 08/Aug/21 Answered by alcoho last updated on 09/Aug/21 $${iymc}\:{queztion} \\ $$$${ur}\:{identity}\:{is}\:{noted}\: \\ $$ Terms of Service Privacy…
Question Number 150006 by mathdanisur last updated on 08/Aug/21 $$\mathrm{If}\:\:\mathrm{x}\:+\:\mathrm{y}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\mathrm{which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{cannot}\:\mathrm{be}\:\:\mathrm{xy}.? \\ $$ Answered by dumitrel last updated on 08/Aug/21 $${xy}={k}\Rightarrow{x}+\frac{{k}}{{x}}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{2}{x}^{\mathrm{2}} −{x}+\mathrm{2}{k}=\mathrm{0}\Rightarrow\bigtriangleup\geqslant\mathrm{0}\Rightarrow \\ $$$$\mathrm{1}−\mathrm{16}{k}\geqslant\mathrm{0}\Rightarrow{k}\in\left(−\infty;\frac{\mathrm{1}}{\mathrm{16}}\right]\Rightarrow{k}\notin\left(\frac{\mathrm{1}}{\mathrm{16}};\infty\right)…
Question Number 150007 by mathdanisur last updated on 08/Aug/21 $$\int\:\:\frac{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}}{\mathrm{x}^{\mathrm{6}} \:+\:\mathrm{1}}\:\mathrm{dx}\:=\:? \\ $$ Answered by Ar Brandon last updated on 08/Aug/21 $${I}=\int\frac{{x}^{\mathrm{2}} +{x}}{{x}^{\mathrm{6}} +\mathrm{1}}{dx}=\int\frac{{x}^{\mathrm{2}}…