Question Number 90097 by jagoll last updated on 21/Apr/20 $$\left(\sqrt{\mathrm{3}+\sqrt{\mathrm{8}}}\right)^{\mathrm{x}} \:+\left(\sqrt{\mathrm{3}−\sqrt{\mathrm{8}}}\right)^{\mathrm{x}} \:=\:\mathrm{6} \\ $$ Commented by john santu last updated on 21/Apr/20 $$\left(\mathrm{3}+\sqrt{\mathrm{8}}\right)\left(\mathrm{3}−\sqrt{\mathrm{8}}\right)=\mathrm{1} \\ $$$$\mathrm{3}−\sqrt{\mathrm{8}}\:=\:\frac{\mathrm{1}}{\mathrm{3}+\sqrt{\mathrm{8}}}\:…
Question Number 90092 by jagoll last updated on 21/Apr/20 $$\mathrm{G}\left(\sqrt{\mathrm{x}+\mathrm{5}}\right)\:=\:\mathrm{x} \\ $$$$\mathrm{G}\left(\mathrm{x}^{\mathrm{2}} \right)\:=\:\mathrm{x}^{\mathrm{a}} −\mathrm{b} \\ $$$$\mathrm{find}\:\mathrm{a}+\mathrm{b}\: \\ $$ Commented by john santu last updated on…
Question Number 155621 by mathdanisur last updated on 02/Oct/21 Answered by ghimisi last updated on 03/Oct/21 $$\frac{{a}+{b}+{c}}{\mathrm{6}}=\mathrm{1}\Rightarrow \\ $$$$\left(\frac{{a}+{d}}{{a}}\right)^{\frac{{a}}{\mathrm{6}}} \left(\frac{{b}+{e}}{{b}}\right)^{\frac{{b}}{\mathrm{6}}} \left(\frac{{c}+{f}}{{c}}\right)^{\frac{{c}}{\mathrm{6}}} \leqslant\frac{{a}}{\mathrm{6}}\centerdot\frac{{a}+{d}}{{a}}+\frac{{b}}{\mathrm{6}}\centerdot\frac{{b}+{e}}{{b}}+\frac{{c}}{\mathrm{6}}\centerdot\frac{{c}+{f}}{{c}}=\mathrm{2}\Rightarrow \\ $$$$\left(\frac{{a}+{d}}{{a}}\right)^{{a}} \left(\frac{{b}+{c}}{{b}}\right)^{{b}}…
Question Number 155620 by mathdanisur last updated on 02/Oct/21 $$\mathrm{Find}:\:\:\:\boldsymbol{\Omega}\:=\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\underset{\boldsymbol{\mathrm{x}}-\mathrm{1}} {\overset{\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} -\boldsymbol{\mathrm{e}}} {\int}}\mathrm{cos}\left(\mathrm{t}^{\mathrm{5}} \right)\mathrm{dt}}{\mathrm{3}^{\boldsymbol{\mathrm{x}}} \:-\:\mathrm{3}}\:=\:? \\ $$ Answered by mindispower last updated on 04/Oct/21…
Question Number 90086 by I want to learn more last updated on 21/Apr/20 Commented by I want to learn more last updated on 21/Apr/20 $$\mathrm{Ohh},\:\:\mathrm{i}\:\mathrm{grab}…
Question Number 24548 by Physics lover last updated on 20/Nov/17 $${prove}\:{that}\: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{r}} {\sum}}\left\{{n}\left({n}−\frac{{r}}{\mathrm{2}}\right)^{\mathrm{2}} \right\}=\:{r}\centerdot\underset{{n}=\mathrm{1}} {\overset{{r}/\mathrm{2}} {\sum}}{n}^{\mathrm{2}} \\ $$$$\:{where}\:\:\:{r}\:=\:\mathrm{2}{k}\:;\:{k}\:\in\:\mathbb{N} \\ $$ Answered by jota…
Question Number 155619 by mathdanisur last updated on 02/Oct/21 $$\boldsymbol{\Omega}\:=\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\mathrm{x}^{\mathrm{49}} }{\mathrm{1}\:+\:\mathrm{x}\:+\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}^{\mathrm{3}} \:…\:\mathrm{x}^{\mathrm{100}} }\:\mathrm{dx}\:=\:? \\ $$ Answered by mindispower last updated on 04/Oct/21…
Question Number 24542 by Tinkutara last updated on 20/Nov/17 $${Prove}\:{that}\:{coefficient}\:{of}\:{x}^{{n}} \:{in} \\ $$$$\frac{{a}+{bx}+{cx}^{\mathrm{2}} }{{e}^{{x}} }\:{is}\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\left[{cn}^{\mathrm{2}} −\left({b}+{c}\right){n}+{a}\right] \\ $$ Commented by Tinkutara last updated on…
Question Number 155618 by mathdanisur last updated on 02/Oct/21 $$\mathrm{if}\:\:\mathrm{x};\mathrm{y}\in\mathbb{R}\:\:\mathrm{and}\:\:\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} =\mathrm{16}\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\mathrm{x}^{\mathrm{4}} \:+\:\mathrm{y}^{\mathrm{4}} \:+\:\mathrm{2x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:\geqslant\:\mathrm{4x}\:+\:\mathrm{36} \\ $$ Terms of Service Privacy Policy…
Question Number 24540 by Tinkutara last updated on 20/Nov/17 $${Prove}\:{that} \\ $$$$\left({i}\right)\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{2}} }{{n}!}=\mathrm{2}{e}. \\ $$$$\left({ii}\right)\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{3}} }{{n}!}=\mathrm{5}{e}. \\ $$$$\left({iii}\right)\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{4}} }{{n}!}=\mathrm{15}{e}.…