Question Number 151042 by mathdanisur last updated on 17/Aug/21 $$\mathrm{if}\:\:\sqrt{\sqrt[{\mathrm{3}}]{\mathrm{9}}\:−\:\mathrm{1}}\:+\:\sqrt{\sqrt[{\mathrm{3}}]{\mathrm{16}}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{4}}}\:=\:\sqrt{\boldsymbol{\mathrm{x}}}\:\:;\:\:\boldsymbol{\mathrm{x}}\in\mathbb{Z} \\ $$$$\mathrm{find}\:\:\boldsymbol{\mathrm{x}}=? \\ $$ Commented by MJS_new last updated on 18/Aug/21 $${x}\notin\mathbb{Z} \\ $$ Terms…
Question Number 151043 by mathdanisur last updated on 17/Aug/21 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{system}: \\ $$$$\begin{cases}{\boldsymbol{\mathrm{y}}\:=\:\frac{\mathrm{2x}}{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\\{\boldsymbol{\mathrm{z}}\:=\:\frac{\mathrm{2y}}{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }}\\{\boldsymbol{\mathrm{x}}\:=\:\frac{\mathrm{2z}}{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }}\end{cases} \\ $$ Answered by john_santu last updated on 18/Aug/21 $$\mathrm{x}=\mathrm{y}=\mathrm{z}\:\Rightarrow\mathrm{x}=\frac{\mathrm{2x}}{\mathrm{1}−\mathrm{x}^{\mathrm{2}}…
Question Number 151035 by text last updated on 17/Aug/21 $$\begin{vmatrix}{}&{}\\{}&{}\end{vmatrix} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 85490 by oustmuchiya@gmail.com last updated on 22/Mar/20 $${Given}\:{that}\:{the}\:{expression}\:\mathrm{2}{x}^{\mathrm{3}} +{px}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{9}\:{is}\:{exactly}\:{divisable}\:{by}\:{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{5},\:{find}\:{the}\:{value}\:{of}\:\boldsymbol{\mathrm{p}}\:{and}\:\boldsymbol{\mathrm{q}}.\:{Hence}\:{factorise}\:{the}\:{expression}\:{fully} \\ $$ Commented by john santu last updated on 22/Mar/20 $${nothing}\:{q}\:{in}\:{this}\:{equation}. \\…
Question Number 151013 by mathdanisur last updated on 17/Aug/21 $$\mathrm{If}\:\:\mathrm{log}\left(\mathrm{log}\boldsymbol{\mathrm{x}}\right)^{\frac{\mathrm{3}\boldsymbol{\mathrm{log}}\left(\boldsymbol{\mathrm{logx}}\right)}{\boldsymbol{\mathrm{log}}\left(\boldsymbol{\mathrm{log}}\left(\boldsymbol{\mathrm{logx}}\right)\right)}\:} \:=\:\mathrm{27} \\ $$$$\mathrm{Find}\:\:\boldsymbol{\mathrm{x}}=? \\ $$ Answered by Olaf_Thorendsen last updated on 17/Aug/21 $$\mathrm{log}\left(\mathrm{log}{x}\right)^{\frac{\mathrm{3log}\left(\mathrm{log}{x}\right)}{\mathrm{log}\left(\mathrm{log}\left(\mathrm{log}{x}\right)\right)}} \:=\:\mathrm{27}\:=\:\mathrm{3}^{\mathrm{3}} \\…
Question Number 19945 by Tinkutara last updated on 18/Aug/17 $$\mathrm{If}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{equation} \\ $$$${x}^{\mathrm{2}} \:+\:{px}\:+\:{q}\:=\:\mathrm{0}\:\mathrm{and}\:\alpha^{\mathrm{2}} ,\:\beta^{\mathrm{2}} \:\mathrm{are}\:\mathrm{roots}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{equation}\:{x}^{\mathrm{2}} \:−\:{rx}\:+\:{s}\:=\:\mathrm{0},\:\mathrm{show} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{equation}\:{x}^{\mathrm{2}} \:−\:\mathrm{4}{qx}\:+\:\mathrm{2}{q}^{\mathrm{2}} \:−\:{r}\:=\:\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{real}\:\mathrm{roots}. \\…
Question Number 151015 by mathdanisur last updated on 17/Aug/21 Commented by JDamian last updated on 17/Aug/21 Instead of repeating several times a day, why don't you try to clarify your question? Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 19936 by ajfour last updated on 18/Aug/17 $${Find}\:{the}\:{pricipal}\:{value}\:{of}\: \\ $$$$\:\:\:\:\:\left(\mathrm{1}−{i}\right)^{\mathrm{1}+{i}} \:. \\ $$ Commented by prof Abdo imad last updated on 22/Jun/18 $$\left(\mathrm{1}−{i}\right)^{\mathrm{1}+{i}}…
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Question Number 150979 by mathdanisur last updated on 17/Aug/21 $$\mathrm{if}\:\:\mathrm{x}=\sqrt[{\mathrm{3}}]{\mathrm{4}}+\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\:\:\mathrm{find}\:\:\frac{\mathrm{3}}{\mathrm{x}}+\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }=? \\ $$$$ \\ $$ Answered by john_santu last updated on 18/Aug/21 $$\:\mathrm{If}\:\mathrm{x}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{4}}+\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\:\mathrm{then}\:\frac{\mathrm{3}}{\mathrm{x}}+\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}}…