Question Number 121218 by ZiYangLee last updated on 06/Nov/20 $$\mathrm{If}\:{x}=\sqrt{\mathrm{1}+\sqrt{\mathrm{2}}},\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }. \\ $$ Answered by liberty last updated on 06/Nov/20 $$\mathrm{x}^{\mathrm{2}} =\mathrm{1}+\sqrt{\mathrm{2}}\:\Rightarrow\mathrm{x}^{\mathrm{4}} =\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}} \\…
Question Number 121219 by ZiYangLee last updated on 06/Nov/20 $$\mathrm{Given}\:\mathrm{that}\:{a},{b}\:\mathrm{and}\:{c}\:\mathrm{are}\:\mathrm{three}\:\mathrm{consecutive} \\ $$$$\mathrm{numbers},\:\mathrm{where}\:{a}>{b}>{c},\:\mathrm{such}\:\mathrm{that}\:\mathrm{its}\:\mathrm{product} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as}\:\mathrm{its}\:\mathrm{sum},\:\mathrm{that}\:\mathrm{is}\:{abc}={a}+{b}+{c}, \\ $$$$\mathrm{how}\:\mathrm{many}\:\mathrm{such}\:\left({a},{b},{c}\right)? \\ $$ Commented by liberty last updated on 06/Nov/20…
Question Number 186750 by pascal889 last updated on 09/Feb/23 Answered by Frix last updated on 09/Feb/23 $$\frac{\mathrm{1}+\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} +\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} }{\mathrm{1}+\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }=\frac{\mathrm{2}{x}^{\mathrm{4}} +\mathrm{3}{x}^{\mathrm{2}} +\frac{\mathrm{9}}{\mathrm{8}}}{\mathrm{2}{x}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{2}}}= \\…
Question Number 186751 by 073 last updated on 09/Feb/23 Answered by Rasheed.Sindhi last updated on 09/Feb/23 $$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c} \\ $$$$\mathrm{f}\left(\mathrm{x}−\mathrm{1}\right)+\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}+\mathrm{1}\right)=\mathrm{x}^{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{f}\left(\mathrm{2}\right)=? \\ $$$$\:…
Question Number 55669 by problem solverd last updated on 01/Mar/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 186739 by Davidtim last updated on 09/Feb/23 $$\sqrt[{\mathrm{3}}]{\mathrm{5}+\sqrt[{\mathrm{3}}]{\mathrm{5}+\sqrt[{\mathrm{3}}]{\mathrm{5}+…}}}=? \\ $$ Answered by pablo1234523 last updated on 09/Feb/23 $${x}=\sqrt[{\mathrm{3}}]{\mathrm{5}+\sqrt[{\mathrm{3}}]{\mathrm{5}+\sqrt[{\mathrm{3}}]{\mathrm{5}+…}}} \\ $$$$\Rightarrow{x}=\sqrt[{\mathrm{3}}]{\mathrm{5}+{x}} \\ $$$$\Rightarrow{x}^{\mathrm{3}} −{x}−\mathrm{5}=\mathrm{0}…
Question Number 55658 by otchereabdullai@gmail.com last updated on 01/Mar/19 $$\mathrm{find}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{following}\:\mathrm{quadratic}\:\mathrm{equation} \\ $$$$\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}\:}\right)\mathrm{x}^{\mathrm{2}} \:+\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\mathrm{x}\:=\mathrm{2} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 01/Mar/19 $$\alpha+\beta=−\frac{\left(\mathrm{1}+\sqrt{\mathrm{2}}\:\right)}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:}\:\:\:\alpha\beta=\frac{−\mathrm{2}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:}…
Question Number 186705 by aba last updated on 08/Feb/23 $$\mathrm{a},\mathrm{b}>\mathrm{0}\:,\:\mathrm{a}+\mathrm{b}=\mathrm{2} \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{a}^{\mathrm{2b}} +\mathrm{b}^{\mathrm{2a}} +\left(\frac{\mathrm{a}−\mathrm{b}}{\mathrm{2}}\right)^{\mathrm{2}} \leqslant\mathrm{2} \\ $$ Commented by aba last updated on 08/Feb/23 help��please…
Question Number 121154 by zakirullah last updated on 05/Nov/20 Answered by Bird last updated on 05/Nov/20 $${let}\:{f}\left({n}\right)={u}_{{n}} \:\Rightarrow{u}_{{n}} =\mathrm{6}{u}_{{m}−\mathrm{1}} −\mathrm{9}{u}_{{n}−\mathrm{2}} \\ $$$${u}_{\mathrm{0}} =\mathrm{1}\:\:{and}\:{u}_{\mathrm{2}} =\mathrm{2} \\…
Question Number 186689 by SANOGO last updated on 08/Feb/23 $$\underset{{n}={o}} {\overset{+{oo}} {\sum}}\:\frac{{x}^{\mathrm{3}{n}} }{\left(\mathrm{2}{n}\right)!}\:\:=\:\:\:? \\ $$ Commented by mr W last updated on 08/Feb/23 $${why}\:{not}\:\infty\:{instead}\:{of}\:{oo}? \\…