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 $${f}\left({x}\right)=\mathrm{2}^{{x}} +\mathrm{3}^{{x}}…
Question Number 182000 by mr W last updated on 03/Dec/22 $${if}\:\boldsymbol{{a}}−\mathrm{2}\boldsymbol{{b}}+\mathrm{3}\boldsymbol{{c}}−\mathrm{4}\boldsymbol{{d}}+\mathrm{5}\boldsymbol{{e}}−\mathrm{6}\boldsymbol{{f}}=\mathrm{0},\:{find} \\ $$$${the}\:{maximum}\:{of} \\ $$$$\frac{\mid\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}+\boldsymbol{{d}}+\boldsymbol{{e}}+\boldsymbol{{f}}\mid}{\:\sqrt{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} +\boldsymbol{{d}}^{\mathrm{2}} +\boldsymbol{{e}}^{\mathrm{2}} +\boldsymbol{{f}}^{\mathrm{2}} }}. \\ $$ Commented by…
Question Number 181998 by Tolmasbek last updated on 03/Dec/22 Answered by a.lgnaoui last updated on 03/Dec/22 $$\alpha=\mathrm{3}\left(\frac{\pi}{\mathrm{18}}+\frac{\pi{n}}{\mathrm{9}}\right) \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{tan}\:\left(\frac{\pi}{\mathrm{18}}+\frac{\pi{n}}{\mathrm{9}}\right)\left[\mathrm{3}−\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{18}}+\frac{\pi{n}}{\mathrm{9}}\right)\right]}{\mathrm{1}−\mathrm{3tan}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{18}}+\frac{\pi{n}}{\mathrm{9}}\right)} \\ $$$$=\frac{\mathrm{tan}\:\left(\frac{\pi}{\mathrm{9}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{n}\right)\left[\mathrm{3}−\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{9}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{n}\right)\right]\right.\right.}{\mathrm{1}−\mathrm{3tan}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{9}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{n}\right)\right.}…
Question Number 50925 by kaivan.ahmadi last updated on 22/Dec/18 $$\mathrm{factor}\:\:\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\mathrm{and} \\ $$$$\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\frac{\mathrm{1}}{\mathrm{3}}\:\:\mathrm{and} \\ $$$$\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} −\mathrm{x}−\frac{\mathrm{1}}{\mathrm{3}} \\ $$ Commented by Abdo…
Question Number 50924 by Saorey last updated on 22/Dec/18 $$\mathrm{factor}\:\mathrm{the}\:\mathrm{expression}: \\ $$$$\mathrm{E}={x}^{\mathrm{5}} +{x}^{\mathrm{4}} +\mathrm{1} \\ $$ Answered by math1967 last updated on 22/Dec/18 $${x}^{\mathrm{5}} −{x}^{\mathrm{2}}…