Question Number 111534 by Aina Samuel Temidayo last updated on 04/Sep/20 $$\mathrm{If}\:\mathrm{x}=\frac{\mathrm{1}+\sqrt{\mathrm{2016}}}{\mathrm{2}},\:\mathrm{then} \\ $$$$\mathrm{4x}^{\mathrm{3}} −\mathrm{2019x}−\mathrm{2017}\:\mathrm{equals}? \\ $$ Answered by Lordose last updated on 04/Sep/20 $$…
Question Number 45982 by MrW3 last updated on 19/Oct/18 $${Find}\:{the}\:{value}\left({s}\right)\:{of}\:{a}\:{such}\:{that} \\ $$$${a}^{{x}} \geqslant{ax}\:{with}\:{a},\:{x}\in{R}. \\ $$ Commented by ajfour last updated on 19/Oct/18 Answered by ajfour…
Question Number 177042 by Ar Brandon last updated on 30/Sep/22 Answered by a.lgnaoui last updated on 30/Sep/22 $${somme}\:{dds}\:{racinesx}_{\mathrm{1}} +{x}_{\mathrm{2}} =−{a}^{\mathrm{2}} \:\:\:\:\:\:{x}_{\mathrm{1}} {x}_{\mathrm{2}} \:=\frac{{a}+\mathrm{3}}{\mathrm{4}{a}^{\mathrm{2}} −\mathrm{19}{a}−\mathrm{5}} \\…
Question Number 177037 by mr W last updated on 30/Sep/22 Answered by Rasheed.Sindhi last updated on 30/Sep/22 $$\begin{cases}{\frac{{x}+{y}+{z}}{{x}\left({y}+{z}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{x}+{y}+{z}=\frac{{x}\left({y}+{z}\right)}{\mathrm{2}}}\\{\frac{{x}+{y}+{z}}{{y}\left({z}+{x}\right)}=\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow{x}+{y}+{z}=\frac{{y}\left({z}+{x}\right)}{\mathrm{3}}}\\{\frac{{x}+{y}+{z}}{{z}\left({x}+{y}\right)}=\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow{x}+{y}+{z}=\frac{{z}\left({x}+{y}\right)}{\mathrm{4}}}\end{cases}\: \\ $$$$\begin{cases}{\frac{{x}\left({y}+{z}\right)}{\mathrm{2}}=\frac{{y}\left({z}+{x}\right)}{\mathrm{3}}\Rightarrow\mathrm{3}{xy}+\mathrm{3}{zx}=\mathrm{2}{yz}+\mathrm{2}{xy}}\\{\frac{{y}\left({z}+{x}\right)}{\mathrm{3}}=\frac{{z}\left({x}+{y}\right)}{\mathrm{4}}\Rightarrow\mathrm{4}{yz}+\mathrm{4}{xy}=\mathrm{3}{zx}+\mathrm{3}{yz}\:\:}\\{\frac{{z}\left({x}+{y}\right)}{\mathrm{4}}=\frac{{x}\left({y}+{z}\right)}{\mathrm{2}}\Rightarrow\mathrm{4}{xy}+\mathrm{4}{zx}=\mathrm{2}{zx}+\mathrm{2}{yz}}\end{cases}\:\: \\ $$$$\Rightarrow\begin{cases}{\mathrm{3}{xy}+\mathrm{3}{zx}=\mathrm{2}{yz}+\mathrm{2}{xy}}\\{\mathrm{4}{yz}+\mathrm{4}{xy}=\mathrm{3}{zx}+\mathrm{3}{yz}}\\{\mathrm{4}{xy}+\mathrm{4}{zx}=\mathrm{2}{zx}+\mathrm{2}{yz}}\end{cases}\: \\ $$$$\Rightarrow\begin{cases}{{xy}−\mathrm{2}{yz}+\mathrm{3}{zx}=\mathrm{0}}\\{\mathrm{4}{xy}+{yz}−\mathrm{3}{zx}=\mathrm{0}}\\{\mathrm{4}{xy}−\mathrm{2}{yz}+\mathrm{2}{zx}=\mathrm{0}}\end{cases}\: \\…
Question Number 177030 by mr W last updated on 01/Oct/22 $${if}\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}}}=\mathrm{2022} \\ $$$${find}\:\left[{x}\right]=? \\ $$ Answered by TheHoneyCat last updated on 30/Sep/22 $$\mathrm{if}\:{x}\:\mathrm{exists},\:\mathrm{squaring}\:\mathrm{the}\:\mathrm{equlity}\:\mathrm{implies}\:\mathrm{that}: \\ $$$${x}+\mathrm{2022}=\left(\mathrm{2022}\right)^{\mathrm{2}}…
Question Number 45936 by Meritguide1234 last updated on 19/Oct/18 Commented by MJS last updated on 19/Oct/18 $$\mathrm{2}.\mathrm{115410982027}… \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 176995 by blackmamba last updated on 29/Sep/22 $$\:\:{x}^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{4}}{x}+\frac{\mathrm{1}}{\mathrm{8}}=\mathrm{0} \\ $$$$\:{x}=? \\ $$ Answered by mr W last updated on 29/Sep/22 $$\Delta=\left(\frac{\mathrm{1}}{\mathrm{16}}\right)^{\mathrm{2}} +\left(−\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{3}}…
Question Number 176988 by cortano1 last updated on 29/Sep/22 $$\:\begin{cases}{{a}^{\mathrm{3}} =\mathrm{3}{ab}^{\mathrm{2}} +\mathrm{11}}\\{{b}^{\mathrm{3}} =\mathrm{3}{a}^{\mathrm{2}} {b}+\mathrm{2}}\end{cases}\:;\:{a},{b}\:\in\mathrm{R} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =? \\ $$ Commented by mr W last…
Question Number 45905 by peter frank last updated on 18/Oct/18 Answered by math1967 last updated on 18/Oct/18 $${let}\:{no}\:{of}\:{boys}={x}\:{and}\:{no}\:{of}\:{girls}={y} \\ $$$$\therefore\frac{\mathrm{50}{x}+\mathrm{54}{y}}{{x}+{y}}=\mathrm{52} \\ $$$$\mathrm{50}{x}+\mathrm{54}{y}=\mathrm{52}{x}+\mathrm{52}{y} \\ $$$$\mathrm{2}{x}=\mathrm{2}{y}\Rightarrow\frac{{x}}{{y}}=\frac{\mathrm{1}}{\mathrm{1}} \\…
Question Number 176968 by Tolmasbek last updated on 28/Sep/22 $$ \\ $$$$\:\:\:{f}\left({x}\right)+\mathrm{3}{f}\left(\frac{\mathrm{1}}{{x}}\right)=\frac{\mathrm{1}}{{x}}\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right)=?\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com