Question Number 160922 by KONE last updated on 09/Dec/21 $${monter}\:{que}\:\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)^{{n}} +\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)^{{n}\:} \\ $$$${est}\:{divisible}\:{par}\:\mathrm{2}^{{n}} \\ $$$${beson}\:{d}'{aide}\:{svp} \\ $$ Answered by mindispower last updated on 09/Dec/21 $${Starte}\:{withe}…
Question Number 160910 by geron last updated on 09/Dec/21 Answered by blackmamba last updated on 09/Dec/21 $$\:\:\:\:\:\:\sqrt[{\mathrm{3}}]{{a}^{\mathrm{2}} }+\sqrt[{\mathrm{3}}]{{b}^{\mathrm{2}} }+\mathrm{1}+\mathrm{2}\sqrt[{\mathrm{3}}]{{ab}}−\mathrm{2}\sqrt[{\mathrm{3}}]{{a}}−\mathrm{2}\sqrt[{\mathrm{3}}]{{b}}\:=\:\mathrm{49}+\mathrm{20}\sqrt[{\mathrm{3}}]{\mathrm{6}} \\ $$$$ \\ $$ Commented by…
Question Number 29837 by abdo imad last updated on 12/Feb/18 $${let}\:{give}\:\:{T}_{{n}} \left({x}\right)={cos}\left({n}\:{arcosx}\right)\:{with}\:{x}\in\left[−\mathrm{1},\mathrm{1}\right] \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:{T}_{{n}} \:{is}\:{a}\:{polynomial}\:{and}\:{T}_{{n}} \in{Z}\left[{x}\right] \\ $$$$\left.\mathrm{2}\right){calculate}\:{T}_{\mathrm{1}} ,\:{T}_{\mathrm{2}} ,\:{T}_{\mathrm{3}} ,{and}\:{T}_{\mathrm{4}} \\ $$$$\left.\mathrm{3}\right)\:{prove}\:{that}\:{T}_{{n}+\mathrm{2}} \left({x}\right)=\mathrm{2}{x}\:{T}_{{n}+\mathrm{1}} \left({x}\right)−{T}_{{n}}…
Question Number 95373 by Don08q last updated on 24/May/20 $$\: \\ $$$$\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{m}\:\mathrm{for}\:\mathrm{which}\:\mathrm{the}\:\mathrm{roots} \\ $$$$\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:{x}^{\mathrm{3}} \:+\:\mathrm{6}{x}^{\mathrm{2}} \:+\:\mathrm{11}{x}\:+{m}\:=\:\mathrm{0} \\ $$$$\:\mathrm{form}\:\mathrm{a}\:\mathrm{linear}\:\mathrm{sequence}. \\ $$$$ \\ $$ Commented by Rasheed.Sindhi…
Question Number 29832 by abdo imad last updated on 12/Feb/18 $${p}\:{is}\:{a}\:{polynomial}\:{having}\:{n}\:{roots}\:{x}_{{i}} \:\:{with}\:{x}_{{i}} \neq{x}_{{j}} \:{for}\:{i}\neq{j} \\ $$$${prove}\:{that}\:\sum_{{i}=\mathrm{1}} ^{{n}} \:\:\frac{{p}^{''} \left({x}_{{i}} \right)}{{p}^{'} \left({x}_{{i}} \right)}=\mathrm{0} \\ $$ Terms…
Question Number 160895 by HongKing last updated on 08/Dec/21 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{x}^{\boldsymbol{\mathrm{n}}} \:+\:\mathrm{x}^{\boldsymbol{\mathrm{n}}-\mathrm{1}} \:+\:…\:+\:\mathrm{1}}\:\mathrm{dx}\:=\:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\:\left[\boldsymbol{\psi}^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{2}}{\mathrm{n}}\right)\:-\:\boldsymbol{\psi}^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{1}}{\mathrm{n}}\right)\right] \\ $$ Answered by Kamel last…
Question Number 160894 by HongKing last updated on 08/Dec/21 Answered by Kamel last updated on 09/Dec/21 $${a}_{{n}+\mathrm{1}} ={a}_{{n}+\mathrm{2}} −{a}_{{n}+\mathrm{3}} +{a}_{{n}} −{a}_{{n}+\mathrm{1}} \\ $$$${a}_{\mathrm{1}} +\underset{{k}=\mathrm{1}} {\overset{\mathrm{2021}}…
Question Number 29820 by Tinkutara last updated on 12/Feb/18 Commented by ajfour last updated on 12/Feb/18 $$\left(\mathrm{1}\right)\:{in}\:{G}.{P}. \\ $$ Commented by math solver last updated…
Question Number 160883 by cortano last updated on 08/Dec/21 $$\:\sqrt{\mathrm{x}+\mathrm{1}}\:=\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{x}−\mathrm{2}\:\sqrt[{\mathrm{3}}]{\mathrm{2x}+\mathrm{1}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2x}+\mathrm{1}}\:−\mathrm{3}\:}\: \\ $$$$\:\mathrm{x}\:\in\mathbb{R}\: \\ $$ Answered by MJS_new last updated on 08/Dec/21 $$\mathrm{this}\:\mathrm{looks}\:\mathrm{suspicious}\:\mathrm{to}\:\mathrm{me}\:\mathrm{because} \\ $$$$\sqrt{\varphi+\mathrm{1}}=\left(\mathrm{2}\varphi+\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}}…
Question Number 160879 by amin96 last updated on 08/Dec/21 Commented by cortano last updated on 08/Dec/21 $$\Leftrightarrow\:\mathrm{27}.\mathrm{x}^{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)} \:=\:\mathrm{x}^{\frac{\mathrm{10}}{\mathrm{3}}} \\ $$$$\Leftrightarrow\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{27}\right)+\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)\right)^{\mathrm{2}} =\:\frac{\mathrm{10}}{\mathrm{3}}\:\mathrm{log}\:_{\mathrm{3}}…