Question Number 20488 by xing last updated on 27/Aug/17 Commented by ajfour last updated on 27/Aug/17 $${Question}\:{is}: \\ $$$${Given}\:\:\:{a}=−\sqrt{\mathrm{99}}+\sqrt{\mathrm{999}}+\sqrt{\mathrm{9999}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{b}=\sqrt{\mathrm{99}}−\sqrt{\mathrm{999}}+\sqrt{\mathrm{9999}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{c}=\sqrt{\mathrm{99}}+\sqrt{\mathrm{999}}−\sqrt{\mathrm{9999}} \\ $$$${Find}\:{the}\:{value}\:{of}…
Question Number 151554 by mathdanisur last updated on 21/Aug/21 Answered by MJS_new last updated on 21/Aug/21 $${f}\:\left({x},\:{y}\right)\:=\frac{{x}\left({x}+{y}\right)}{\mathrm{4}{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} } \\ $$$$\frac{{d}}{{dy}}\left[{f}\:\left({x},\:{y}\right)\right]=\mathrm{0} \\ $$$$\frac{{x}\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{xy}−{y}^{\mathrm{2}} \right)}{\left(\mathrm{4}{x}^{\mathrm{2}}…
Question Number 151549 by mathdanisur last updated on 21/Aug/21 $$\mathrm{Compare}: \\ $$$$\mathrm{2}^{\mathrm{2}^{\mathrm{2}^{.^{.^{.} } } } } \:\:\:\:\:\mathrm{and}\:\:\:\:\:\mathrm{3}^{\mathrm{3}^{\mathrm{3}^{.^{.^{.} } } } } \\ $$$$\mathrm{Here}\:\mathrm{it}\:\mathrm{is}\:\mathrm{raised}\:\mathrm{1001}\:\mathrm{times}\:\mathrm{a}\:\mathrm{square}, \\ $$$$\mathrm{1000}\:\mathrm{times}\:\mathrm{a}\:\mathrm{cube}.…
Question Number 86009 by M±th+et£s last updated on 26/Mar/20 $${solve}\:{in}\:{R}\::\left[\frac{{x}}{\mathrm{2}}\right]+\left[\frac{\mathrm{2}{x}}{\mathrm{3}}\right]−{x}=\mathrm{0} \\ $$ Answered by MJS last updated on 26/Mar/20 $$\left[\frac{{x}}{\mathrm{2}}\right]\in\mathbb{Z}\wedge\left[\frac{\mathrm{2}{x}}{\mathrm{3}}\right]\in\mathbb{Z}\Rightarrow{x}\in\mathbb{Z} \\ $$$$\mathrm{let}\:{k},\:{m},\:{n}\:\in\mathbb{Z} \\ $$$$\left[\frac{{x}}{\mathrm{2}}\right]=\begin{cases}{\frac{{x}}{\mathrm{2}};\:{x}=\mathrm{2}{m}}\\{\frac{{x}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}};\:{x}=\mathrm{2}{m}+\mathrm{1}}\end{cases} \\…
Question Number 86000 by oustmuchiya@gmail.com last updated on 26/Mar/20 $${solve}\:{the}\:{equation}\:\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{4} \\ $$ Answered by TANMAY PANACEA. last updated on 26/Mar/20 $$\left({x}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{3}} =\mathrm{4}^{\mathrm{3}} \\…
Question Number 151533 by mathdanisur last updated on 21/Aug/21 $$\mathrm{Compare}: \\ $$$$\mathrm{2}^{\mathrm{2}^{\mathrm{2}} } \:\:\:\mathrm{and}\:\:\:\mathrm{3}^{\mathrm{3}^{\mathrm{3}} } \\ $$ Commented by Rasheed.Sindhi last updated on 21/Aug/21 $$\mathrm{3}^{\mathrm{3}^{\mathrm{3}}…
Question Number 151531 by bekzodjumayev last updated on 21/Aug/21 Commented by bekzodjumayev last updated on 21/Aug/21 $${Please}\:{help} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 20430 by Tinkutara last updated on 26/Aug/17 $$\mathrm{Let}\:{a},\:{b}\:\mathrm{and}\:{c}\:\mathrm{be}\:\mathrm{such}\:\mathrm{that}\:{a}\:+\:{b}\:+\:{c}\:=\:\mathrm{0} \\ $$$$\mathrm{and} \\ $$$${P}\:=\:\frac{{a}^{\mathrm{2}} }{\mathrm{2}{a}^{\mathrm{2}} \:+\:{bc}}\:+\:\frac{{b}^{\mathrm{2}} }{\mathrm{2}{b}^{\mathrm{2}} \:+\:{ca}}\:+\:\frac{{c}^{\mathrm{2}} }{\mathrm{2}{c}^{\mathrm{2}} \:+\:{ab}} \\ $$$$\mathrm{is}\:\mathrm{defined}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{P}? \\ $$ Commented…
Question Number 151490 by mathdanisur last updated on 21/Aug/21 $$\begin{cases}{\mathrm{a}^{\mathrm{2}} =\mathrm{2a}+\mathrm{b}}\\{\mathrm{b}^{\mathrm{2}} =\mathrm{a}+\mathrm{2b}}\end{cases}\:\:\mathrm{and}\:\:\mathrm{a}\neq\mathrm{b}\:\:\mathrm{find}\:\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{1}} \\ $$ Answered by iloveisrael last updated on 21/Aug/21 $$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}}…
Question Number 151481 by mathdanisur last updated on 21/Aug/21 $$\mathrm{if}\:\:\mathrm{x};\mathrm{y};\mathrm{z};\mathrm{t}\in\mathbb{R}\:\:\mathrm{and} \\ $$$$\mathrm{x}+\mathrm{y}+\mathrm{z}\:\leqslant\:\mathrm{3t}\:\:;\:\:\mathrm{y}+\mathrm{z}+\mathrm{t}\:\leqslant\:\mathrm{3x} \\ $$$$\mathrm{z}+\mathrm{t}+\mathrm{x}\:\leqslant\:\mathrm{3y}\:\:;\:\:\mathrm{t}+\mathrm{x}+\mathrm{y}\:\leqslant\:\mathrm{3z} \\ $$$$\mathrm{compare}:\:\:\mathrm{x};\mathrm{y};\mathrm{z};\mathrm{t} \\ $$ Commented by dumitrel last updated on 21/Aug/21…