Question Number 158383 by HongKing last updated on 03/Nov/21 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{tan}^{\mathrm{4}} \left(\boldsymbol{\mathrm{x}}\right)}\:+\:\frac{\mathrm{1}}{\mathrm{10}}\:=\:\frac{\mathrm{2}}{\mathrm{1}\:+\:\mathrm{3}\:\mathrm{tan}^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}\right)} \\ $$$$ \\ $$ Answered by MJS_new last updated on 03/Nov/21…
Question Number 158378 by Rasheed.Sindhi last updated on 03/Nov/21 $${How}\:{to}\:{graph}\:{order}\:{pair}\:\left(\mathrm{3}+\mathrm{5}{i}\:,\:\mathrm{4}−\mathrm{2}{i}\right)? \\ $$ Answered by MJS_new last updated on 03/Nov/21 $$\mathrm{you}\:\mathrm{would}\:\mathrm{need}\:\mathrm{a}\:\mathrm{4}−\mathrm{dimensional}\:\mathrm{coordinate} \\ $$$$\mathrm{system}… \\ $$ Commented…
Question Number 27300 by mondodotto@gmail.com last updated on 04/Jan/18 Commented by mondodotto@gmail.com last updated on 04/Jan/18 $$\mathrm{any}\:\mathrm{one}\:\mathrm{to}\:\mathrm{help}\:\mathrm{this}\:\mathrm{please} \\ $$ Answered by bsayani309@gmail.com last updated on…
Question Number 92839 by i jagooll last updated on 09/May/20 $$\mathrm{let}\:\mathrm{a}\:\mathrm{is}\:\mathrm{complex}\:\mathrm{number}\:\mathrm{such}\: \\ $$$$\mathrm{that}\:\mathrm{a}^{\mathrm{10}} \:+\:\mathrm{a}^{\mathrm{5}} \:+\mathrm{1}\:=\:\mathrm{0}. \\ $$$$\mathrm{find}\:\mathrm{a}^{\mathrm{2005}} \:+\:\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2005}} }\:? \\ $$ Answered by john santu…
Question Number 92835 by unknown last updated on 09/May/20 Commented by unknown last updated on 09/May/20 $$\mathrm{let}\:{a},{b},{c}\:\mathrm{is}\:\mathrm{real}\:\mathrm{positive}\:\mathrm{number}.\:{a}+{b}+{c}=\mathrm{3} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:: \\ $$$${A}=\frac{\mathrm{2}−{a}^{\mathrm{3}} }{{a}}+\frac{\mathrm{2}−{b}^{\mathrm{3}} }{{b}}+\frac{\mathrm{2}−{c}^{\mathrm{3}} }{{c}} \\…
Question Number 158364 by HongKing last updated on 03/Nov/21 $$\mathrm{if}\:\:\mathrm{a};\mathrm{b};\mathrm{c}>\mathrm{0}\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\mathrm{a}^{\mathrm{2}\boldsymbol{\mathrm{a}}-\left(\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\right)} \:\centerdot\:\mathrm{b}^{\mathrm{2}\boldsymbol{\mathrm{b}}-\left(\boldsymbol{\mathrm{c}}+\boldsymbol{\mathrm{a}}\right)} \:\centerdot\:\mathrm{c}^{\mathrm{2}\boldsymbol{\mathrm{c}}-\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\right)} \:\geqslant\:\mathrm{1} \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact:…
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Question Number 92831 by unknown last updated on 09/May/20 Commented by unknown last updated on 09/May/20 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}\:\left({p},{q},{r}\right)\:\mathrm{if}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{are}\:\mathrm{real}\:\mathrm{number} \\ $$ Commented by unknown last updated on…
Question Number 158363 by HongKing last updated on 03/Nov/21 $$\mathrm{x};\mathrm{y};\mathrm{z}>\mathrm{0} \\ $$$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\begin{cases}{\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:+\:\mathrm{z}^{\mathrm{3}} \:+\:\mathrm{3}\centerdot\left(\sqrt[{\mathrm{3}}]{\mathrm{x}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{y}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{z}}\right)\:=\:\mathrm{12}}\\{\mathrm{x}\centerdot\mathrm{y}\centerdot\mathrm{z}\:=\:\mathrm{1}}\end{cases} \\ $$$$ \\ $$ Answered by GuruBelakangPadang last…
Question Number 158366 by HongKing last updated on 03/Nov/21 Answered by MathsFan last updated on 03/Nov/21 $$\frac{\mathrm{lne}}{\mathrm{lnx}\bullet\mathrm{lnx}}+\frac{\mathrm{lne}}{\left(\mathrm{lne}−\mathrm{lnx}\right)\left(\mathrm{lne}−\mathrm{lnx}\right)}=\mathrm{8} \\ $$$$\mathrm{say}\:\:\mathrm{a}=\mathrm{lnx} \\ $$$$\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2a}+\mathrm{a}^{\mathrm{2}} }=\mathrm{8} \\ $$$$\:\mathrm{2a}^{\mathrm{2}}…