Question Number 152063 by Tawa11 last updated on 25/Aug/21 $$\mathrm{If}\:\:\mathrm{x}\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{real}\:\mathrm{number}\:\mathrm{and}\:\:\:\mathrm{y}\:\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\:\:\:\:\frac{\mathrm{x}^{\mathrm{2}} \:\:+\:\:\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:\:+\:\:\mathrm{x}\:\:+\:\:\mathrm{1}}, \\ $$$$\mathrm{show}\:\mathrm{that}\:\:\:\:\:\mid\mathrm{y}\:\:\:−\:\:\:\frac{\mathrm{4}}{\mathrm{3}}\mid\:\:\:\leqslant\:\:\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$ Answered by mr W last updated on 25/Aug/21 $${y}=\frac{{x}^{\mathrm{2}}…
Question Number 20983 by Tinkutara last updated on 09/Sep/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ordered}\:\mathrm{triples} \\ $$$$\left({a},\:{b},\:{c}\right)\:\mathrm{of}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{such}\:\mathrm{that} \\ $$$${abc}\:=\:\mathrm{108}. \\ $$ Answered by dioph last updated on 10/Sep/17 $$\mathrm{108}\:=\:\mathrm{2}^{\mathrm{2}} ×\mathrm{3}^{\mathrm{3}}…
Question Number 152049 by mathdanisur last updated on 25/Aug/21 $$\mathrm{If}\:\:\mathrm{a};\mathrm{b}\geqslant\mathrm{1}\:\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\left(\mathrm{a}+\mathrm{1}+\frac{\mathrm{a}+\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }\right)^{\boldsymbol{\mathrm{a}}} \centerdot\:\left(\mathrm{b}+\mathrm{1}+\frac{\mathrm{b}+\mathrm{1}}{\mathrm{b}^{\mathrm{2}} }\right)^{\boldsymbol{\mathrm{b}}} \geqslant\:\mathrm{2}^{\mathrm{2}\left(\mathrm{1}+\sqrt{\boldsymbol{\mathrm{ab}}}\right)} \: \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 152052 by john_santu last updated on 25/Aug/21 $$\:\:\mathrm{knowns}\:\mathrm{x}_{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{x}_{\mathrm{1}} }\:,\:\mathrm{x}_{\mathrm{3}} =\frac{\mathrm{3}}{\mathrm{x}_{\mathrm{2}} }\:,\:\mathrm{x}_{\mathrm{4}} =\frac{\mathrm{4}}{\mathrm{x}_{\mathrm{3}} } \\ $$$$,\:\mathrm{x}_{\mathrm{5}} =\frac{\mathrm{5}}{\mathrm{x}_{\mathrm{4}} }\:,\:…,\:\mathrm{x}_{\mathrm{8}} =\frac{\mathrm{8}}{\mathrm{x}_{\mathrm{7}} }.\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{x}_{\mathrm{1}} ×\mathrm{x}_{\mathrm{2}}…
Question Number 152032 by mathdanisur last updated on 25/Aug/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 152034 by mathdanisur last updated on 25/Aug/21 $$\begin{cases}{\mathrm{3x}^{\mathrm{2}} \:-\:\mathrm{2y}\:=\:-\:\frac{\mathrm{17}}{\mathrm{3}}}\\{\mathrm{y}^{\mathrm{2}} \:-\:\mathrm{6x}\:=\:\mathrm{7}}\end{cases}\:\:\:\Rightarrow\:\:\mathrm{xy}\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 25/Aug/21 $$\begin{cases}{\mathrm{3x}^{\mathrm{2}} \:-\:\mathrm{2y}\:=\:-\:\frac{\mathrm{17}}{\mathrm{3}}}\\{\mathrm{y}^{\mathrm{2}} \:-\:\mathrm{6x}\:=\:\mathrm{7}}\end{cases}\:\:\:\Rightarrow\:\:\mathrm{xy}\:=\:?\: \\…
Question Number 152030 by mathdanisur last updated on 25/Aug/21 Commented by mathdanisur last updated on 26/Aug/21 $$\boldsymbol{\mathrm{S}}\mathrm{er},\:\Omega_{{n}} +\Omega_{{n}+\mathrm{2}} +…+\Omega_{{n}×\mathrm{6}} =\mathrm{1} \\ $$ Commented by ghimisi…
Question Number 152029 by mathdanisur last updated on 25/Aug/21 Commented by ghimisi last updated on 25/Aug/21 $$\left(\mathrm{1}−\mathrm{4}{x}\right)^{\mathrm{2}} \:\:{or}\:\:\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} \\ $$ Commented by mathdanisur last updated…
Question Number 152019 by mathdanisur last updated on 25/Aug/21 $$\Omega\:=\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:{Li}_{\mathrm{2}} \left({x}\right)\:{log}\left(\mathrm{1}+{x}\right)\:{dx}\:=\:? \\ $$$${Li}_{\mathrm{2}} \left({x}\right)−{polylogaritm}\:{function} \\ $$ Answered by mnjuly1970 last updated on 25/Aug/21…
Question Number 20935 by Tinkutara last updated on 08/Sep/17 $$\mathrm{If}\:\mid{z}\:+\:\omega\mid^{\mathrm{2}} \:=\:\mid{z}\mid^{\mathrm{2}} \:+\:\mid\omega\mid^{\mathrm{2}} ,\:\mathrm{where}\:{z}\:\mathrm{and}\:\omega \\ $$$$\mathrm{are}\:\mathrm{complex}\:\mathrm{numbers},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\frac{{z}}{\omega}\:\mathrm{is}\:\mathrm{purely}\:\mathrm{real} \\ $$$$\left(\mathrm{2}\right)\:\frac{{z}}{\omega}\:\mathrm{is}\:\mathrm{purely}\:\mathrm{imaginary} \\ $$$$\left(\mathrm{3}\right)\:{z}\bar {\omega}\:+\:\bar {{z}}\omega\:=\:\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{amp}\left(\frac{{z}}{\omega}\right)\:=\:\frac{\pi}{\mathrm{2}}…