Question Number 32372 by jasno91 last updated on 24/Mar/18 Answered by $@ty@m last updated on 24/Mar/18 $$\mathrm{3}.\:{D}=\frac{\mathrm{147}}{\mathrm{4}}+\frac{\mathrm{202}}{\mathrm{5}}+\frac{\mathrm{77}}{\mathrm{2}}=\frac{\mathrm{2313}}{\mathrm{20}}=\mathrm{115}\frac{\mathrm{13}}{\mathrm{20}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 163437 by cortano1 last updated on 07/Jan/22 Commented by mr W last updated on 07/Jan/22 $${m}^{\mathrm{3}} +{n}^{\mathrm{3}} +{p}^{\mathrm{3}} =\left(\sqrt[{\mathrm{3}}]{\mathrm{13}}\right)^{\mathrm{3}} +\left(\sqrt[{\mathrm{3}}]{\mathrm{53}}\right)^{\mathrm{3}} +\left(\sqrt[{\mathrm{3}}]{\mathrm{103}}\right)^{\mathrm{3}} −\mathrm{3}×\frac{\mathrm{1}}{\mathrm{3}} \\…
Question Number 163433 by HongKing last updated on 06/Jan/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 163434 by HongKing last updated on 06/Jan/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 163430 by amin96 last updated on 06/Jan/22 Answered by qaz last updated on 07/Jan/22 $$\frac{\mathrm{A}}{\mathrm{B}}=\frac{\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{999}} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2n}−\mathrm{1}\right)\left(\mathrm{2n}\right)}}{\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{999}} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{999}+\mathrm{n}\right)\left(\mathrm{1999}−\mathrm{n}\right)}} \\ $$$$=\frac{\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{999}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2n}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2n}}\right)=\mathrm{H}_{\mathrm{1998}}…
Question Number 32333 by abdo imad last updated on 23/Mar/18 $${decompose}\:{F}\left({x}\right)\:=\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)}\:{inside}\:{R}\left({x}\right). \\ $$ Commented by abdo imad last updated on 01/Apr/18 $${F}\left({x}\right)=\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}\:=\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}}…
Question Number 32334 by abdo imad last updated on 23/Mar/18 $${p}\:{is}\:{apolynomial}\:{having}\:{n}\:{roots}\:\:\left({x}_{{i}} \right)\:{with}\:{x}_{{i}} \neq{xj}\:{for} \\ $$$${i}\neq{j}\:\:{calculate}\:\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{\mathrm{1}}{\mathrm{1}−{x}_{{k}} }\:. \\ $$ Commented by abdo imad last…
Question Number 32332 by abdo imad last updated on 23/Mar/18 $${calculate}\:\:\sum_{{p}=\mathrm{1}} ^{{n}} \:\:\:\frac{{p}}{\mathrm{1}+{p}^{\mathrm{2}} \:+{p}^{\mathrm{4}} }\:\:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 32331 by abdo imad last updated on 23/Mar/18 $${p}\:{is}\:{a}\:{polynomial}\:{having}\:{n}\:{simples}\:{roots}\:\left({x}_{{i}} \right)_{\mathrm{1}\leqslant{i}\leqslant{n}} \\ $$$${prove}\:{that}\:\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{p}^{'} \left({x}_{{k}} \right)}\:=\mathrm{0} \\ $$ Terms of Service Privacy Policy…
Question Number 32330 by abdo imad last updated on 23/Mar/18 $${let}\:{p}_{{n}} \left({x}\right)=\left({x}+\mathrm{1}\right)^{\mathrm{6}{n}+\mathrm{1}} \:−{x}^{\mathrm{6}{n}+\mathrm{1}} \:−\mathrm{1}\:{with}\:{n}\:{integr} \\ $$$${prove}\:{that}\:\forall{n}\:\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} \:{divide}\:{p}_{{n}} \left({x}\right). \\ $$ Commented by abdo imad…