Question Number 31494 by abdo imad last updated on 09/Mar/18 $${prove}\:{that}\:{x}^{\mathrm{2}} \:{divide}\:\left({x}+\mathrm{1}\right)^{\underset{} {{n}}} \:−{nx}−\mathrm{1}\:.{nintegr}. \\ $$ Commented by Rasheed.Sindhi last updated on 10/Mar/18 $$\:\:\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{n}} −\mathrm{nx}−\mathrm{1}…
Question Number 31492 by abdo imad last updated on 09/Mar/18 $${find}\:{all}\:{polynomial}\:{p}\left({x}\right)\:{wich}\:{verify}\: \\ $$$$\forall{k}\in{Z}\:\:\:\int_{{k}} ^{{k}+\mathrm{1}} {p}\left({x}\right){dx}={k}+\mathrm{1}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 31490 by abdo imad last updated on 09/Mar/18 $${simplify}\:{p}\left({x}\right)=\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)….\left(\mathrm{1}+{x}^{\mathrm{2}{n}} \right)\:{with}\:{n}\:{fromN} \\ $$$${then}\:{find}\:{the}\:{roots}\:{of}\:{p}\left({x}\right). \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 31491 by abdo imad last updated on 09/Mar/18 $${let}\:{p}\left({x}\right)=\:{x}^{{n}} \:+{a}_{{n}−\mathrm{1}} {x}^{{n}−\mathrm{1}} \:+….\:{a}_{\mathrm{1}} {x}\:+{a}_{{o}} \\ $$$${if}\:\:\xi\:\:{is}\:{roots}\:{of}\:{p}\left({x}\right)\:{prove}\:{that}\:\mid\xi\mid\:\leqslant\:\mathrm{1}+{max}_{\mathrm{0}\leqslant{i}\leqslant{n}−\mathrm{1}} \:\mid{a}_{{i}} \mid \\ $$ Terms of Service Privacy…
Question Number 162552 by amin96 last updated on 30/Dec/21 $$\boldsymbol{\alpha}_{\mathrm{1}} <\boldsymbol{\alpha}_{\mathrm{2}} <\boldsymbol{\alpha}_{\mathrm{3}} <\ldots<\boldsymbol{\alpha}_{{k}} \\ $$$$\frac{\mathrm{2}^{\mathrm{289}} +\mathrm{1}}{\mathrm{2}^{\mathrm{17}} +\mathrm{1}}=\mathrm{2}^{\boldsymbol{\alpha}_{\mathrm{1}} } +\mathrm{2}^{\boldsymbol{\alpha}_{\mathrm{2}} } +\ldots+\mathrm{2}^{\boldsymbol{\alpha}_{{k}} } \:\:\:\:\:\:\:\boldsymbol{\mathrm{k}}=? \\ $$$$…
Question Number 31485 by mondodotto@gmail.com last updated on 09/Mar/18 Answered by ajfour last updated on 09/Mar/18 $$\left({i}\right)\:\:\:{let}\:\:\:\:{y}−\mathrm{2}\:=\:{t}\left({x}−\mathrm{3}\right) \\ $$$$\Rightarrow\:\:\:\:\frac{{dy}}{{dx}}=\left({x}−\mathrm{3}\right)\frac{{dt}}{{dx}}+{t} \\ $$$${Then}\:\:{diff}.\:{eq}.\:{becomes} \\ $$$$\:\:\:\:\:\:\:\:\left({x}−\mathrm{3}\right)\frac{{dt}}{{dx}}+{t}\:=\:\frac{{t}\left({x}−\mathrm{3}\right)}{\left({t}+\mathrm{1}\right)\left({x}−\mathrm{3}\right)} \\ $$$$\Rightarrow\:\:\left({x}−\mathrm{3}\right)\frac{{dt}}{{dx}}\:+{t}\:=\frac{{t}}{{t}+\mathrm{1}}…
Question Number 97019 by bshahid010@gmail.com last updated on 06/Jun/20 Commented by PRITHWISH SEN 2 last updated on 06/Jun/20 $$\left(\mathrm{2}+\mathrm{2}.\mathrm{8}+\mathrm{2}.\mathrm{8}^{\mathrm{2}} +….+\mathrm{2}.\mathrm{8}^{\mathrm{29}} \right)+\left(\mathrm{4}+\mathrm{4}.\mathrm{8}+\mathrm{4}.\mathrm{8}^{\mathrm{2}} +…+\mathrm{4}.\mathrm{8}^{\mathrm{29}} \right) \\ $$$$−\left(\mathrm{8}+\mathrm{8}.\mathrm{8}+\mathrm{8}.\mathrm{8}^{\mathrm{2}}…
Question Number 162533 by mnjuly1970 last updated on 30/Dec/21 Answered by Ar Brandon last updated on 30/Dec/21 $${x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{2}=\mathrm{0} \\ $$$$\alpha+\beta=−\mathrm{2} \\ $$$$\alpha\beta=−\mathrm{2} \\ $$$$\frac{\alpha^{\mathrm{4}}…
Question Number 31456 by rahul 19 last updated on 08/Mar/18 $$\mathbb{F}{ind}\:{sum}\:{of} \\ $$$${S}=\:\frac{\mathrm{2}}{\mathrm{3}}\:+\:\frac{\mathrm{4}}{\mathrm{3}^{\mathrm{2}} }\:+\:\frac{\mathrm{6}}{\mathrm{3}^{\mathrm{3}} }\:+\:\frac{\mathrm{8}}{\mathrm{3}^{\mathrm{4}} }\:+……+\infty\:? \\ $$ Commented by MJS last updated on 08/Mar/18…
Question Number 162521 by HongKing last updated on 30/Dec/21 $$\mathrm{For}\:\mathrm{every}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{number}\:\boldsymbol{\mathrm{x}}\:,\:\mathrm{let} \\ $$$$\mathrm{g}\left(\mathrm{x}\right)\:=\underset{\boldsymbol{\mathrm{r}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\left(\mathrm{x}+\mathrm{1}\right)^{\boldsymbol{\mathrm{r}}+\mathrm{1}} \:-\:\mathrm{x}^{\boldsymbol{\mathrm{r}}+\mathrm{1}} \right)^{\frac{\mathrm{1}}{\boldsymbol{\mathrm{r}}}} \\ $$$$\mathrm{Find}:\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{g}\left(\mathrm{x}\right)}{\mathrm{x}} \\ $$ Terms of Service Privacy Policy…