Question Number 84915 by M±th+et£s last updated on 17/Mar/20 $${if}\: \\ $$$${x}>\mathrm{0},{y}>\mathrm{0},{z}>\mathrm{0} \\ $$$${show}\:{that} \\ $$$$\frac{{x}+{y}}{{z}}+\frac{{z}+{y}}{\:{x}}+\frac{{z}+{x}}{{y}}\geqslant\mathrm{6}\:\: \\ $$ Answered by TANMAY PANACEA last updated on…
Question Number 84913 by M±th+et£s last updated on 17/Mar/20 $$ \\ $$$${sin}\frac{\pi}{\mathrm{14}}\:{sin}\frac{\mathrm{3}\pi}{\mathrm{14}}\:{sin}\frac{\mathrm{5}\pi}{\mathrm{15}}=? \\ $$ Commented by jagoll last updated on 17/Mar/20 $$\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{14}}\:?\:\mathrm{or}\:\mathrm{not} \\ $$ Commented…
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Question Number 150435 by mathdanisur last updated on 12/Aug/21 $$\boldsymbol{\mathrm{S}}\mathrm{olve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\mid\mathrm{x}\:-\:\mathrm{3}\mid^{\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:-\:\mathrm{8x}\:+\:\mathrm{15}}{\boldsymbol{\mathrm{x}}\:-\:\mathrm{2}}} \:=\:\mathrm{1} \\ $$ Answered by amin96 last updated on 12/Aug/21 $${x}\neq\mathrm{2}\:\:\:\Rightarrow\:\:{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{15}=\mathrm{0}\:\:\left({x}−\mathrm{5}\right)\left({x}−\mathrm{3}\right)=\mathrm{0}…
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Question Number 19350 by Tinkutara last updated on 10/Aug/17 $$\mathrm{Prove}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \:+\:{z}_{\mathrm{2}} \mid\:=\:\mid{z}_{\mathrm{1}} \:−\:{z}_{\mathrm{2}} \mid\:\Leftrightarrow \\ $$$$\mathrm{arg}\left({z}_{\mathrm{1}} \right)\:−\:\mathrm{arg}\left({z}_{\mathrm{2}} \right)\:=\:\frac{\pi}{\mathrm{2}} \\ $$ Answered by ajfour last updated…
Question Number 19351 by Tinkutara last updated on 10/Aug/17 $$\mathrm{Prove}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \:+\:{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:=\:\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} \:+\:\mid{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:\Leftrightarrow \\ $$$$\frac{{z}_{\mathrm{1}} }{{z}_{\mathrm{2}} }\:\mathrm{is}\:\mathrm{purely}\:\mathrm{imaginary}\:\mathrm{number}. \\ $$ Commented by…
Question Number 19349 by Tinkutara last updated on 10/Aug/17 $$\mathrm{Prove}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \:+\:{z}_{\mathrm{2}} \:+\:{z}_{\mathrm{3}} \:+\:….\:+\:{z}_{{n}} \mid\:\leqslant \\ $$$$\mid{z}_{\mathrm{1}} \mid\:+\:\mid{z}_{\mathrm{2}} \mid\:+\:\mid{z}_{\mathrm{3}} \mid\:+\:….\:+\:\mid{z}_{{n}} \mid \\ $$ Commented by Tinkutara…
Question Number 84871 by tw000001 last updated on 17/Mar/20 $$\mathrm{If}\:\mathrm{you}\:\mathrm{know} \\ $$$$\left(\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}\right)^{\mathrm{2}} +\left(\frac{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ca}}\right)^{\mathrm{2}} +\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}\right)^{\mathrm{2}} =\mathrm{3}, \\…