Question Number 194732 by MathsFan last updated on 14/Jul/23 Commented by York12 last updated on 15/Jul/23 $${do}\:{not}\:{participate}\:{IYMC} \\ $$$${it}\:{is}\:{fake} \\ $$ Answered by mahdipoor last…
Question Number 194735 by sonukgindia last updated on 14/Jul/23 Answered by TheHoneyCat last updated on 14/Jul/23 $$\mathrm{If}\:{x}\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant},\:\left({i}.{e}.\:{x}\left(\mathrm{2}\right)={x}×\mathrm{2}\:\right) \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{is}: \\ $$$${x}\left(\mathrm{2}\right)=\frac{\mathrm{4}\sqrt{\mathrm{7}}−\mathrm{2}\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{7}}+\sqrt{\mathrm{5}}} \\ $$$$ \\ $$$$\mathrm{But}\:\mathrm{I}'\mathrm{m}\:\mathrm{guessing}\:{x}\:\mathrm{is}\:\mathrm{here}\:\mathrm{a}\:\mathrm{function}……
Question Number 194685 by cortano12 last updated on 13/Jul/23 Answered by qaz last updated on 14/Jul/23 $$\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} ={e}^{\frac{\mathrm{1}}{{x}}{ln}\left(\mathrm{1}+{x}\right)} ={e}^{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{2}} +…} \\ $$$$={e}\left(\mathrm{1}+\left(−\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{2}} +…\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{2}} +…\right)^{\mathrm{2}} +…\right.…
Question Number 194700 by MathsFan last updated on 13/Jul/23 Commented by MathsFan last updated on 13/Jul/23 $${help}\:{please} \\ $$ Commented by Tinku Tara last updated…
Question Number 194697 by cortano12 last updated on 13/Jul/23 $$\:\:\:\:\underbrace{\:} \\ $$ Answered by MM42 last updated on 13/Jul/23 $$\frac{{tanx}−{tan}\mathrm{3}{x}}{{tanx}}=\mathrm{3}\Rightarrow\frac{{tan}\mathrm{3}{x}}{{tanx}}=−\mathrm{2} \\ $$$$\frac{{cotx}}{{cotx}+{cot}\mathrm{3}{x}}=\frac{{tan}\mathrm{3}{x}}{{tanx}+{tan}\mathrm{3}{x}} \\ $$$$=\frac{−\mathrm{2}}{\mathrm{1}−\mathrm{2}}=\mathrm{2}\:\checkmark \\…
Question Number 194709 by MM42 last updated on 13/Jul/23 $${Show}\:{that}\:\:{in}\:{fibonacci}\:{sequence} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{f}_{\mathrm{3}{n}} ={f}_{{n}} ^{\mathrm{3}} +{f}_{{n}+\mathrm{1}} ^{\mathrm{3}} −{f}_{{n}−\mathrm{1}} ^{\mathrm{3}} \\ $$$$ \\ $$ Answered by TheHoneyCat…
Question Number 194693 by MM42 last updated on 13/Jul/23 $${if}\:\:\:{f}_{{n}} ={f}_{{n}−\mathrm{1}} +{f}_{{n}−\mathrm{2}} \:\:;\:\:{f}_{\mathrm{1}} ={f}_{\mathrm{2}} =\mathrm{1} \\ $$$${then}\:\:\:{prove}\:{that}\:\:\:\mathrm{5}\mid{f}_{\mathrm{5}{n}} \:\: \\ $$ Answered by Frix last updated…
Question Number 194695 by horsebrand11 last updated on 13/Jul/23 $$\:\:\:\:\:\downharpoonleft\underline{\:} \\ $$ Answered by som(math1967) last updated on 13/Jul/23 $$\boldsymbol{{let}}\:\frac{\boldsymbol{{x}}}{\boldsymbol{{a}}+\boldsymbol{{b}}−\boldsymbol{{c}}}=\frac{\boldsymbol{{y}}}{\boldsymbol{{b}}+\boldsymbol{{c}}−\boldsymbol{{a}}}=\frac{\boldsymbol{{z}}}{\boldsymbol{{c}}+\boldsymbol{{a}}−\boldsymbol{{b}}}=\boldsymbol{{k}} \\ $$$$\boldsymbol{{x}}=\boldsymbol{{k}}\left(\boldsymbol{{a}}+\boldsymbol{{b}}−\boldsymbol{{c}}\right) \\ $$$$\boldsymbol{{y}}=\boldsymbol{{k}}\left(\boldsymbol{{b}}+\boldsymbol{{c}}−\boldsymbol{{a}}\right) \\…
Question Number 194689 by cortano12 last updated on 13/Jul/23 Answered by MM42 last updated on 13/Jul/23 $${f}_{{i}} :\mathrm{1},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{8},\mathrm{13},\mathrm{21},\mathrm{34},\mathrm{55},\mathrm{89},… \\ $$$$\Rightarrow\forall\:{n}=\mathrm{5}{k}\:\:;\:\mathrm{5}\mid{f}_{{n}} \:\:\:!!?? \\ $$$$\Rightarrow{if}\:\:{n}<\mathrm{10000}\:\rightarrow\:{ans}=\mathrm{4999}\:\checkmark \\ $$$${if}\:\:{f}_{{n}}…
Question Number 194688 by CrispyXYZ last updated on 13/Jul/23 $$\mathrm{find}\:\mathrm{all}\:\mathrm{function}\:{f}:\:\mathbb{R}\:\rightarrow\:\mathbb{R}\:\mathrm{such}\:\mathrm{that}\:\forall{x},\:{y}\in\mathbb{R}, \\ $$$${f}\left({x}−{f}\left({y}\right)\right)={f}\left({f}\left({y}\right)\right)+{xf}\left({y}\right)+{f}\left({x}\right)−\mathrm{1}. \\ $$ Answered by Tinku Tara last updated on 13/Jul/23 $${put}\:{x}={f}\left({y}\right) \\ $$$${f}\left(\mathrm{0}\right)={f}\left({x}\right)+{x}^{\mathrm{2}}…