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 $$\mathrm{If}\:\mathrm{you}\:\mathrm{know} \\ $$$$\left(\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}\right)^{\mathrm{2}} +\left(\frac{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ca}}\right)^{\mathrm{2}} +\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}\right)^{\mathrm{2}} =\mathrm{3}, \\…
Question Number 150404 by mathdanisur last updated on 12/Aug/21 $$\Omega\:=\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{\mathrm{log}\left(\mathrm{x}\right)}{\left(\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\mathrm{dx}\:=\:? \\ $$ Answered by Ar Brandon last updated on 12/Aug/21 $$\Omega\left({a}\right)=\int_{\mathrm{0}}…
Question Number 19332 by Tinkutara last updated on 09/Aug/17 $$\mathrm{Let}\:{S}_{{n}} \:=\:{n}^{\mathrm{2}} \:+\:\mathrm{20}{n}\:+\:\mathrm{12},\:{n}\:\mathrm{a}\:\mathrm{positive} \\ $$$$\mathrm{integer}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{possible} \\ $$$$\mathrm{values}\:\mathrm{of}\:{n}\:\mathrm{for}\:\mathrm{which}\:{S}_{{n}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{perfect} \\ $$$$\mathrm{square}? \\ $$ Commented by mrW1 last…
Question Number 150402 by mathdanisur last updated on 12/Aug/21 $$\mathrm{For}\:\:\mathrm{m}\geqslant\mathrm{1} \\ $$$$\boldsymbol{\Omega}\:=\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{\mathrm{x}\:\mathrm{ln}^{\boldsymbol{\mathrm{m}}} \:\left(\mathrm{x}\right)}{\mathrm{e}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{1}}\:=\:\mathrm{2}\:\mathrm{ln}^{\boldsymbol{\mathrm{m}}} \:\boldsymbol{\zeta}\left(\mathrm{3}\right) \\ $$ Commented by yadiirama last updated on…