Question Number 97303 by bobhans last updated on 07/Jun/20 $$\boldsymbol{\mathrm{G}}\mathrm{iven}\:\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} +\mathrm{x}_{\mathrm{3}} \:=\:\mathrm{0}\:,\:\mathrm{y}_{\mathrm{1}} \:+\:\mathrm{y}_{\mathrm{2}} +\mathrm{y}_{\mathrm{3}} \:=\:\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{x}_{\mathrm{1}} \mathrm{y}_{\mathrm{1}} +\:\mathrm{x}_{\mathrm{2}} \mathrm{y}_{\mathrm{2}} \:+\:\mathrm{x}_{\mathrm{3}} \mathrm{y}_{\mathrm{3}} \:=\:\mathrm{0}\:.\:\mathrm{The}\:\mathrm{value} \\…
Question Number 31763 by RAMANUJAN last updated on 14/Mar/18 $${please}\:{find}\:{the}\:{integral}\:{solutions}\:\left({x}\:{and}\:{y}\right)\: \\ $$$$\left({xy}−\mathrm{7}\right)^{\mathrm{2}} \:={x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \\ $$ Answered by MJS last updated on 14/Mar/18 $$\left(−\mathrm{7},\mathrm{0}\right) \\…
Question Number 162819 by mnjuly1970 last updated on 01/Jan/22 $$ \\ $$$$\:\:\:\:\:#{combinatorial}\:{mathematics}# \\ $$$$\:\:\:\:\:\:\mathrm{I}{n}\:{how}\:{many}\:\:{subsets}\:{of}\:\mathrm{10}\: \\ $$$${members}\:{of}\:\:{the}\:{set}\:,\:\left\{\:\mathrm{1},\:\mathrm{2},\:…,\:\mathrm{20}\:\right\} \\ $$$$\:\:\:{is}\:{there}\:{no}\:{difference}\:{between} \\ $$$$\:{two}\:{members}\:{of}\:\:\:\:\mathrm{5}\:?\: \\ $$$${a}:\:\:\mathrm{2}^{\:\mathrm{10}} \:\:\:\:\:\:{b}\::\:\:\mathrm{3}^{\:\mathrm{5}} \:\:\:\:\:\:\:{c}\::\:\:\:\mathrm{2}^{\:\mathrm{8}} \:\:\:\:\:\:\:\:{d}:\:\:\:\mathrm{10}^{\:\mathrm{4}}…
Question Number 162814 by amin96 last updated on 01/Jan/22 Answered by MathsFan last updated on 01/Jan/22 $$\:{let}\:\:{log}_{\mathrm{3}} {x}={u}\:\rightarrow\:{x}=\mathrm{3}^{{u}} \\ $$$$\:\mathrm{2}^{{u}} =\sqrt{\mathrm{3}^{{u}} }+\mathrm{1} \\ $$$$\:\mathrm{2}^{{u}} −\sqrt{\mathrm{3}^{{u}}…
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Question Number 162792 by HongKing last updated on 01/Jan/22 $$\mathrm{Find}: \\ $$$$\boldsymbol{\Omega}\:\:=\:\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{\mathrm{arctan}\left(\mathrm{x}\right)}{\mathrm{x}\centerdot\left(\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{x}\:+\:\mathrm{1}\right)}\:\mathrm{dx} \\ $$ Answered by phanphuoc last updated on 01/Jan/22 $${we}\:{have}\:{arctan}\left({x}\right)={arctan}\left({tx}\right)\mid_{\mathrm{0}}…
Question Number 31726 by rahul 19 last updated on 13/Mar/18 $${Let}\:{a}_{\mathrm{1}} =\:\frac{\mathrm{1}}{\mathrm{2}}\:,\:{a}_{{k}+\mathrm{1}} ={a}_{{k}} ^{\mathrm{2}} +{a}_{{k}} \forall\:{k}\geqslant\:\mathrm{1}. \\ $$$${then}\:{a}_{\mathrm{101}} \:\:{is}\:{greater}\:{than} \\ $$$$\left.{a}\right)\:\mathrm{1}\: \\ $$$$\left.{b}\right)\:\mathrm{2} \\ $$$$\left.{c}\right)\:\mathrm{3}…
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Question Number 162791 by mnjuly1970 last updated on 01/Jan/22 Answered by amin96 last updated on 01/Jan/22 $$\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{x}}−\mathrm{1}=\mathrm{0}\:\:\:\:\Leftrightarrow{Viets}\:\boldsymbol{\alpha\beta}=−\mathrm{1} \\ $$$$\boldsymbol{\alpha}=−\mathrm{1}+\sqrt{\mathrm{2}}\:\:\:\:\:\boldsymbol{\beta}=−\mathrm{1}−\sqrt{\mathrm{2}} \\ $$$$\boldsymbol{\alpha}^{\mathrm{4}} \boldsymbol{\beta}^{\mathrm{9}} +\mathrm{29}\boldsymbol{\alpha}+\mathrm{58}=\boldsymbol{\beta}^{\mathrm{5}} \left(\boldsymbol{\alpha\beta}\right)^{\mathrm{4}}…
Question Number 162788 by HongKing last updated on 01/Jan/22 $$\mathrm{let}\:\:\mathrm{x};\mathrm{y};\mathrm{z}\:>\:\mathrm{0} \\ $$$$\mathrm{such}\:\mathrm{that}\:\:\mathrm{x}^{\mathrm{4}} +\mathrm{y}^{\mathrm{4}} +\mathrm{z}^{\mathrm{4}} \:=\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{expression}: \\ $$$$\mathrm{P}\:=\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{y}}\:+\:\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{z}}\:+\:\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{x}} \\…