Question Number 164124 by HongKing last updated on 14/Jan/22 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\frac{\mathrm{16}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{4}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{1}^{\boldsymbol{\mathrm{x}}} }{\mathrm{4}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{1}^{\boldsymbol{\mathrm{x}}} }\:=\:\frac{\mathrm{8}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{1}}{\mathrm{65}} \\ $$ Answered by MJS_new last…
Question Number 164116 by mathlove last updated on 14/Jan/22 $$\left({a}+{b}+{c}+{x}\right)^{\mathrm{100}} \\ $$$$\mathrm{50}^{{th}} \:{limit}\:{is}\:{equl}\:{to}=? \\ $$ Commented by mr W last updated on 14/Jan/22 $${what}\:{is}\:{a}\:{first}\:{limit}? \\…
Question Number 98570 by HamraboyevFarruxjon last updated on 14/Jun/20 $$\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}}>\mathrm{0}\:\:\:\:\:\:\:\boldsymbol{{prove}}: \\ $$$$\frac{\boldsymbol{{a}}}{\:\sqrt{\boldsymbol{{a}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{{bc}}}}+\frac{\boldsymbol{{b}}}{\:\sqrt{\boldsymbol{{b}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{{ac}}}}+\frac{\boldsymbol{{c}}}{\:\sqrt{\boldsymbol{{c}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{{ab}}}}\geqslant\mathrm{1} \\ $$$$\boldsymbol{{help}}\:\boldsymbol{{please}}… \\ $$ Commented by MJS last updated on…
Question Number 33032 by rahul 19 last updated on 09/Apr/18 $${f}:{N}\rightarrow{R} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{2005}. \\ $$$${and}\: \\ $$$${f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)+……+{f}\left({n}\right)=\:{n}^{\mathrm{2}} \:{f}\left({n}\right),{n}>\mathrm{1}. \\ $$$${Then}\:{f}\left(\mathrm{2004}\right)=? \\ $$ Commented by Joel578…
Question Number 164077 by mathlove last updated on 13/Jan/22 Answered by cortano1 last updated on 13/Jan/22 $$\:\left(\mathrm{1}\right){x}+{y}+{xy}+\mathrm{1}=\mathrm{4}\Rightarrow\left({x}+\mathrm{1}\right)\left({y}+\mathrm{1}\right)=\mathrm{4} \\ $$$$\left(\mathrm{2}\right){y}+{z}+{yz}+\mathrm{1}=\mathrm{6}\Rightarrow\left({y}+\mathrm{1}\right)\left({z}+\mathrm{1}\right)=\mathrm{6} \\ $$$$\left(\mathrm{3}\right)\:{z}+{x}+{zx}+\mathrm{1}=\mathrm{8}\Rightarrow\left({z}+\mathrm{1}\right)\left({x}+\mathrm{1}\right)=\mathrm{8} \\ $$$$\left(\mathrm{1}\right)×\left(\mathrm{2}\right)×\left(\mathrm{3}\right)\Rightarrow\left({x}+\mathrm{1}\right)\left({y}+\mathrm{1}\right)\left({z}+\mathrm{1}\right)=\pm\sqrt{\mathrm{4}×\mathrm{4}×\mathrm{4}×\mathrm{3}}=\pm\mathrm{8}\sqrt{\mathrm{3}} \\ $$$$\:{z}+\mathrm{1}=\pm\mathrm{2}\sqrt{\mathrm{3}}…
Question Number 164068 by HongKing last updated on 13/Jan/22 Answered by Rasheed.Sindhi last updated on 13/Jan/22 $$\mathrm{sin}\left(\mathrm{x}\right)=\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{8}}}\:\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{x}=\frac{\pi}{\mathrm{6}}+\mathrm{2}{k}\pi\:;\:\forall{k}\in\mathbb{Z} \\ $$ Commented by mr W…
Question Number 164071 by kdaramaths last updated on 13/Jan/22 Answered by Ar Brandon last updated on 13/Jan/22 $${Z}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{e}^{{ikx}} =\frac{\mathrm{1}−{e}^{{i}\left({n}+\mathrm{1}\right){x}} }{\mathrm{1}−{e}^{{ix}} } \\ $$$$\:\:\:=\frac{{e}^{{i}\left(\frac{\left({n}+\mathrm{1}\right){x}}{\mathrm{2}}\right)}…
Question Number 98521 by pranesh last updated on 14/Jun/20 Answered by abdomathmax last updated on 15/Jun/20 $$\mathrm{we}\:\mathrm{have}\:\mid\mathrm{x}\mid+\mathrm{x}\:\geqslant\mathrm{0}\:\:\mathrm{for}\:\mathrm{all}\:\mathrm{x} \\ $$$$\mathrm{if}\:\:\mathrm{x}\:\geqslant\mathrm{0}\:\mathrm{we}\:\mathrm{get}\:\mathrm{x}+\mathrm{x}\geqslant\mathrm{0}\:\mathrm{true} \\ $$$$\mathrm{if}\:\mathrm{x}\:\leqslant\mathrm{0}\:\mathrm{we}\:\mathrm{get}\:−\mathrm{x}+\mathrm{x}\geqslant\mathrm{0}\:\mathrm{also}\:\mathrm{true}\:\:\mathrm{so} \\ $$$$\mathrm{x}\in\mathrm{D}_{\mathrm{f}} \:\Leftrightarrow\:\mathrm{x}>\mathrm{0}\:\mathrm{and}\:\:\mathrm{x}\neq\frac{\pi}{\mathrm{2}}\:+\mathrm{k}\pi\:\:\left(\mathrm{k}\:\in\mathrm{Z}\right) \\…
Question Number 164050 by mathlove last updated on 13/Jan/22 Answered by Ar Brandon last updated on 13/Jan/22 $${x}−\frac{\mathrm{1}}{\mathrm{8}}>\mathrm{0}\neq\mathrm{1}\Rightarrow{x}>\frac{\mathrm{1}}{\mathrm{8}}\neq\frac{\mathrm{9}}{\mathrm{8}}….\left(\mathrm{1}\right) \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{9}}>\mathrm{0}\:\Rightarrow{x}<−\frac{\mathrm{1}}{\mathrm{3}}\cup{x}>\frac{\mathrm{1}}{\mathrm{3}}\:…\left(\mathrm{2}\right) \\ $$$$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{2}−{x}\right)>\mathrm{0}\:\Rightarrow\mathrm{2}−{x}>\mathrm{0}\Rightarrow{x}<\mathrm{2}…\left(\mathrm{3}\right) \\…
Question Number 32977 by Mr eaay last updated on 08/Apr/18 $${Prove}\:{that}\:\:^{{n}} {C}_{{r}} \:\:+\:^{{n}} {C}_{{r}+\mathrm{1}} \:=\:^{{n}+\mathrm{1}} {C}_{{r}+\mathrm{1}} \\ $$$$ \\ $$ Answered by math1967 last updated…