Question Number 121918 by greg_ed last updated on 12/Nov/20 $${a},\:{b}\:{and}\:{c}\:{are}\:{solutions}\:{of}\:{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{8}=\mathrm{0}. \\ $$$${Without}\:{determinating}\:{a},\:{b}\:{and}\:{c}\:;\: \\ $$$${calculate}\:{a}+{b}+{c}. \\ $$ Commented by mr W last updated on…
Question Number 121919 by talminator2856791 last updated on 12/Nov/20 $$\: \\ $$$$\: \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\underset{{k}={p}} {\overset{\infty} {\sum}}\:\mathrm{4}\centerdot\mathrm{3}^{\mathrm{2}−{k}} \:=\:\frac{\mathrm{2}}{\mathrm{9}} \\ $$$$\:\:\:\:\:\:\:\:\:\:{p}\:=\:? \\ $$$$\: \\ $$$$\:…
Question Number 121913 by Jamshidbek2311 last updated on 12/Nov/20 $$\begin{cases}{{x}^{\mathrm{2}} +{y}=\mathrm{36}}\\{{x}+{y}^{\mathrm{2}} =\mathrm{25}\:\:{x}=?\:{y}=?}\end{cases} \\ $$ Answered by MJS_new last updated on 12/Nov/20 $$\left(\mathrm{1}\right)\:{y}=\mathrm{36}−{x}^{\mathrm{2}} \\ $$$$\Rightarrow \\…
Question Number 187440 by mathlove last updated on 17/Feb/23 $${if}\:\:{y}^{{x}−{y}} =\frac{{x}^{\mathrm{2}} }{{y}}\:{then}\:{x}−{y}=? \\ $$ Commented by mr W last updated on 17/Feb/23 $${you}\:{should}\:{know}\:{that}\:{no}\:{solution} \\ $$$${is}\:{possible}\:{with}\:{given}\:{condition}.…
Question Number 187442 by Shrinava last updated on 17/Feb/23 $$\mathrm{If}\:\:\:\:\:\frac{\mathrm{7}^{\boldsymbol{\mathrm{x}}} }{\mathrm{1}\:+\:\mathrm{7}^{\boldsymbol{\mathrm{x}}} }\:\:=\:\:\frac{\mathrm{3}}{\mathrm{7}}\:\:\:\:\:\mathrm{find}\:\:\:\:\frac{\mathrm{2401}^{\boldsymbol{\mathrm{x}}} }{\mathrm{1}\:+\:\mathrm{2401}^{\boldsymbol{\mathrm{x}}} }\:\:=\:\:? \\ $$ Answered by Rasheed.Sindhi last updated on 17/Feb/23 $$\frac{\mathrm{7}^{\boldsymbol{{x}}} }{\mathrm{1}\:+\:\mathrm{7}^{\boldsymbol{{x}}}…
Question Number 187437 by Shrinava last updated on 17/Feb/23 Answered by witcher3 last updated on 17/Feb/23 $$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{k}}\end{pmatrix}\frac{\mathrm{1}}{\mathrm{m}^{\mathrm{k}} }=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{m}}\right)^{\mathrm{n}} −\mathrm{1} \\ $$$$\Leftrightarrow\mathrm{31}+\underset{\mathrm{m}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\mathrm{m}^{\mathrm{n}}…
Question Number 121890 by oustmuchiya@gmail.com last updated on 12/Nov/20 Commented by liberty last updated on 12/Nov/20 $$\left(\mathrm{v}\right)\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{−\mathrm{1}} \right)^{\mathrm{12}} \:=\:\underset{\mathrm{r}\:=\:\mathrm{0}} {\overset{\mathrm{12}} {\sum}}\begin{pmatrix}{\mathrm{12}}\\{\:\:\mathrm{r}}\end{pmatrix}\:\left(\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{12}−\mathrm{r}} .\left(\mathrm{x}^{−\mathrm{1}} \right)^{\mathrm{r}}…
Question Number 56356 by Tawa1 last updated on 15/Mar/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}\:\mathrm{F}\left(\mathrm{x}\right)\:=\:\:\mathrm{log}_{\mathrm{e}} \mathrm{x}\:\:−\:\:\mathrm{x}\:\:\:\mathrm{for}\:\:\:\mathrm{x}\:>\:\mathrm{0}. \\ $$$$\:\:\mathrm{Hence}\:\mathrm{show}\:\mathrm{that}:\:\:\:\mathrm{log}_{\mathrm{e}} \mathrm{x}\:\:\leqslant\:\:\mathrm{x}\:−\:\mathrm{1}.\:\:\mathrm{For}\:\mathrm{all}\:\:\:\mathrm{x}\:>\:\mathrm{0} \\ $$ Commented by Tawa1 last updated on 15/Mar/19 Answered by…
Question Number 121889 by oustmuchiya@gmail.com last updated on 12/Nov/20 Answered by Dwaipayan Shikari last updated on 12/Nov/20 $$\left(\mathrm{1}+{ax}\right)^{{b}} =\mathrm{1}+\mathrm{48}{x}+\mathrm{1008}{x}^{\mathrm{2}} +… \\ $$$${ab}=\mathrm{48} \\ $$$${a}^{\mathrm{2}} \frac{{b}\left({b}−\mathrm{1}\right)}{\mathrm{2}}=\mathrm{1008}…
Question Number 56335 by Tawa1 last updated on 14/Mar/19 $$\mathrm{If}\:\:\:\:\:\:\mathrm{b}^{\mathrm{x}} \:\:=\:\:\left(\frac{\mathrm{b}}{\mathrm{k}}\right)^{\mathrm{y}} \:\:=\:\:\mathrm{k}^{\mathrm{m}} \:\:\:\:\:\:\:\:\:\mathrm{and}\:\:\:\:\:\:\mathrm{b}\:\neq\:\mathrm{1} \\ $$$$\mathrm{show}\:\mathrm{that}:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{x}}\:=\:\frac{\mathrm{1}}{\mathrm{y}}\:=\:\frac{\mathrm{1}}{\mathrm{m}} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 14/Mar/19 $${b}^{{x}}…