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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}. \\ $$$${then}\:{a}_{\mathrm{101}} \:\:{is}\:{greater}\:{than} \\ $$$$\left.{a}\right)\:\mathrm{1}\: \\ $$$$\left.{b}\right)\:\mathrm{2} \\ $$$$\left.{c}\right)\:\mathrm{3}…
Question Number 162794 by HongKing last updated on 01/Jan/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 162791 by mnjuly1970 last updated on 01/Jan/22 Answered by amin96 last updated on 01/Jan/22 $$\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} \\ $$$$\mathrm{such}\:\mathrm{that}\:\:\mathrm{x}^{\mathrm{4}} +\mathrm{y}^{\mathrm{4}} +\mathrm{z}^{\mathrm{4}} \:=\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{expression}: \\ $$$$\mathrm{P}\:=\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{y}}\:+\:\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{z}}\:+\:\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{x}} \\…
Question Number 162785 by john_santu last updated on 01/Jan/22 Answered by Rasheed.Sindhi last updated on 02/Jan/22 $$ \\ $$$$\frac{{x}^{\mathrm{4}} +\mathrm{2}{x}+\mathrm{5}}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)}=\left({x}−\mathrm{1}\right)+\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)} \\ $$$$\frac{{x}^{\mathrm{2}}…
Question Number 162787 by amin96 last updated on 01/Jan/22 $$\boldsymbol{\mathrm{happy}}\:\boldsymbol{\mathrm{new}}\:\boldsymbol{\mathrm{year}} \\ $$$$\left\{\boldsymbol{{a}};\boldsymbol{{b}};\boldsymbol{{c}}\right\}\in\mathbb{Z}−\left\{\mathrm{0}\right\} \\ $$$$\boldsymbol{{p}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{ax}}^{\mathrm{2}} +\boldsymbol{{bx}}+\boldsymbol{{c}}\:\:\:\: \\ $$$$\boldsymbol{{p}}\left(\boldsymbol{{a}}\right)=\mathrm{0} \\ $$$$\boldsymbol{{p}}\left(\boldsymbol{{b}}\right)=\mathrm{0} \\ $$$$\boldsymbol{{p}}\left(\mathrm{1}\right)=? \\ $$ Answered by…
Question Number 162783 by HongKing last updated on 01/Jan/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 162782 by HongKing last updated on 01/Jan/22 Terms of Service Privacy Policy Contact: info@tinkutara.com