Question Number 182073 by Acem last updated on 04/Dec/22 $${Find}\:{the}\:{sum}\:{of}\:{the}\:{solutions}\:{of}\:{the}\:{equation}: \\ $$$$\:\mid\sqrt{{x}}\:−\:\mathrm{2}\mid+\:\sqrt{{x}}\:\left(\sqrt{{x}}\:−\:\mathrm{4}\right)+\:\mathrm{2}=\:\mathrm{0}\:\:\:;\:{x}>\:\mathrm{0} \\ $$ Answered by HeferH last updated on 04/Dec/22 $$\:\mid\sqrt{{x}}\:−\:\mathrm{2}\mid\:=\:\mathrm{4}\sqrt{{x}}\:−\:{x}\:−\:\mathrm{2} \\ $$$$\:\:\sqrt{{x}}\:−\:\mathrm{2}\:=\:\pm\:\left(\mathrm{4}\sqrt{{x}}\:−\:{x}\:−\:\mathrm{2}\right) \\…
Question Number 182074 by Acem last updated on 04/Dec/22 $${Let}\:{x}+\:{xy}+\:{y}=\:\mathrm{54}\:\:\:;\:{x},\:{y}\in\:\mathbb{N}\:,\:{Find}\:{x}+\:{y} \\ $$ Answered by HeferH last updated on 04/Dec/22 $${x}\:+\:{xy}\:+\:{y}\:=\:\mathrm{54} \\ $$$$\:{x}\left(\mathrm{1}\:+\:{y}\right)\:+\:{y}\:=\:\mathrm{54} \\ $$$$\:{x}\left(\mathrm{1}\:+\:{y}\right)\:+\:\left(\mathrm{1}\:+\:{y}\right)\centerdot\mathrm{1}\:=\:\mathrm{55} \\…
Question Number 116529 by mr W last updated on 04/Oct/20 Commented by MJS_new last updated on 04/Oct/20 $$\frac{\mathrm{5}}{\mathrm{197}} \\ $$ Answered by maths mind last…
Question Number 116509 by Khalmohmmad last updated on 04/Oct/20 $$\mathrm{x}^{\mathrm{4}} −\mathrm{48x}^{\mathrm{2}} +\mathrm{x}+\mathrm{565}=\mathrm{0}\: \\ $$$$\mathrm{x}=? \\ $$ Answered by TANMAY PANACEA last updated on 04/Oct/20 $${f}\left({x}\right)={x}^{\mathrm{4}}…
Question Number 116507 by bemath last updated on 04/Oct/20 Answered by bobhans last updated on 04/Oct/20 $$\mathrm{let}\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{2}}}\:=\:\mathrm{r}\:\Rightarrow\:\frac{\mathrm{1}}{\frac{\mathrm{2}+\mathrm{x}}{\mathrm{2x}}}=\:\mathrm{r} \\ $$$$\Rightarrow\frac{\mathrm{2x}}{\mathrm{2}+\mathrm{x}}\:=\:\mathrm{r}\:.\:\mathrm{the}\:\mathrm{equation}\:\mathrm{equivalent} \\ $$$$\mathrm{to}\:\frac{\mathrm{1}}{\left[\frac{\mathrm{1}}{\mathrm{r}+\mathrm{r}}+\frac{\mathrm{1}}{\mathrm{r}+\mathrm{r}}\right]}\:=\:\frac{\mathrm{x}}{\mathrm{36}}\:\Rightarrow\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{\mathrm{2r}}+\frac{\mathrm{1}}{\mathrm{2r}}\right)}\:=\:\frac{\mathrm{x}}{\mathrm{36}} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{\mathrm{r}}\right)}\:=\:\frac{\mathrm{x}}{\mathrm{36}}\:;\:\mathrm{r}\:=\:\frac{\mathrm{x}}{\mathrm{36}} \\ $$$$\Rightarrow\frac{\mathrm{2x}}{\mathrm{2}+\mathrm{x}}\:=\:\frac{\mathrm{x}}{\mathrm{36}}\:;\:\mathrm{x}\neq\mathrm{0}…
Question Number 50972 by Necxx last updated on 22/Dec/18 Answered by mr W last updated on 22/Dec/18 $$\mathrm{2}^{\mathrm{2}{x}} =\mathrm{8}{x} \\ $$$${e}^{\mathrm{2}{x}\mathrm{ln}\:\mathrm{2}} =\mathrm{8}{x} \\ $$$$\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}}{e}^{\mathrm{2}{x}\mathrm{ln}\:\mathrm{2}} =\mathrm{2}{x}\mathrm{ln}\:\mathrm{2}…
Question Number 182043 by depressiveshrek last updated on 03/Dec/22 $${For}\:{a},\:{b},\:{c},\:{d}\:\in\:\mathbb{R} \\ $$$${a}+{b}+{c}+{d}=\mathrm{0} \\ $$$${ab},\:{ac},\:{ad},\:{bc},\:{bd},\:{cd}\:\neq\mathrm{0} \\ $$$${Prove}\:{the}\:{inequality}: \\ $$$$\frac{{ab}}{\left({a}+{b}\right)^{\mathrm{2}} }+\frac{{ac}}{\left({a}+{c}\right)^{\mathrm{2}} }+\frac{{ad}}{\left({a}+{d}\right)^{\mathrm{2}} }+\frac{{bc}}{\left({b}+{c}\right)^{\mathrm{2}} }+\frac{{bd}}{\left({b}+{d}\right)^{\mathrm{2}} }+\frac{{cd}}{\left({c}+{d}\right)^{\mathrm{2}} }\leq−\frac{\mathrm{3}}{\mathrm{2}} \\…
Question Number 182012 by CrispyXYZ last updated on 03/Dec/22 $${f}\left({x}\right)=\mathrm{9}^{{x}} −{m}\centerdot\mathrm{3}^{{x}} +{m}+\mathrm{6} \\ $$$$\exists{x}\in\mathbb{R},\:{f}\left({x}\right)+{f}\left(−{x}\right)=\mathrm{0} \\ $$$$\mathrm{find}\:\mathrm{the}\:\:\mathrm{range}\:\mathrm{of}\:{m}. \\ $$ Answered by mr W last updated on…
Question Number 50932 by Tawa1 last updated on 22/Dec/18 Answered by tanmay.chaudhury50@gmail.com last updated on 22/Dec/18 $${log}_{\left({log}_{{a}} {c}\right)^{\mathrm{2}} } {log}_{{b}} {a}=\frac{−\mathrm{3}}{\mathrm{2}} \\ $$$$\left[\left({log}_{{a}} {c}\right)^{\mathrm{2}} \right]^{\frac{−\mathrm{3}}{\mathrm{2}}}…
Question Number 182004 by CrispyXYZ last updated on 03/Dec/22 $${f}\left({x}\right)=\mathrm{2}^{{x}} +\mathrm{3}^{{x}} −\mathrm{6}^{{x}} \\ $$$$\mathrm{Find}\:{f}\left({x}\right)_{\mathrm{max}} \\ $$ Answered by ARUNG_Brandon_MBU last updated on 03/Dec/22 $${f}\left({x}\right)=\mathrm{2}^{{x}} +\mathrm{3}^{{x}}…