Question Number 177130 by mr W last updated on 01/Oct/22 $${if}\:{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} =\mathrm{5},\:{find}\:{the}\:{minimum} \\ $$$${and}\:{maximum}\:{of}\:{x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} . \\ $$ Commented by Frix last updated on…
Question Number 111583 by Rio Michael last updated on 04/Sep/20 $$\mathrm{without}\:\mathrm{using}\:\mathrm{a}\:\mathrm{substitution}\:\mathrm{for}\:\mathrm{3}^{{x}} \:\mathrm{solve}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\:\:\mathrm{9}^{{x}} −\mathrm{3}^{{x}\:+\:\mathrm{2}} −\mathrm{36}\:=\:\mathrm{0} \\ $$ Answered by bemath last updated on 04/Sep/20…
Question Number 177112 by mr W last updated on 01/Oct/22 $${solve}\:{for}\:\mathbb{R} \\ $$$${a}+{bcd}=\mathrm{100} \\ $$$${b}+{cda}=\mathrm{100} \\ $$$${c}+{dab}=\mathrm{100} \\ $$$${d}+{abc}=\mathrm{100} \\ $$ Answered by aleks041103 last…
Question Number 111560 by dw last updated on 04/Sep/20 Commented by dw last updated on 04/Sep/20 $${Step}\:{by}\:{step}\:{solution} \\ $$ Answered by Her_Majesty last updated on…
Question Number 177083 by mr W last updated on 30/Sep/22 $${x},\:{y}\:{are}\:{different}\:{positive}\:{integers} \\ $$$${with}\:\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}=\frac{\mathrm{1}}{\mathrm{5}}.\:{find}\:{x}+{y}=? \\ $$ Answered by mr W last updated on 30/Sep/22 $${xy}=\mathrm{5}\left({x}+{y}\right) \\…
Question Number 111534 by Aina Samuel Temidayo last updated on 04/Sep/20 $$\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}}…