Question Number 150059 by mathdanisur last updated on 09/Aug/21 $$\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{n}\centerdot\left(\mathrm{2n}\:+\:\mathrm{1}\right)}\:=\:? \\ $$ Answered by Kamel last updated on 09/Aug/21 $${S}=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{n}}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right)=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{+\infty}…
Question Number 84515 by M±th+et£s last updated on 13/Mar/20 $${Q}.{solve} \\ $$$${x}^{\mathrm{3}} −{x}={x}! \\ $$ Answered by mr W last updated on 13/Mar/20 $$\left({x}+\mathrm{1}\right){x}\left({x}−\mathrm{1}\right)={x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)! \\…
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Question Number 150053 by mathdanisur last updated on 09/Aug/21 $$\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…