Question Number 202500 by Calculusboy last updated on 28/Dec/23 Answered by Frix last updated on 28/Dec/23 $$\mathrm{Obviously}\:{x}=\mathrm{1} \\ $$$$\sqrt[{\mathrm{2015}}]{\mathrm{1}+\mathrm{3}−\mathrm{3}}+\sqrt[{\mathrm{2015}}]{−\mathrm{1}−\mathrm{3}+\mathrm{5}}=\mathrm{1}+\mathrm{1}=\mathrm{2} \\ $$ Answered by Rasheed.Sindhi last…
Question Number 202497 by MATHEMATICSAM last updated on 28/Dec/23 $$\mathrm{If}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of} \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{2}{px}\:+\:{q}\:=\:\mathrm{0}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\: \\ $$$$\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{2}{qx}\:+\:{p}\:=\:\mathrm{0}\:\left[{p}\:\neq\:{q}\right]\:\mathrm{then}\:\mathrm{show} \\ $$$$\mathrm{that}\:{p}\:+\:{q}\:+\:\mathrm{1}\:=\:\mathrm{0}. \\ $$ Answered by BaliramKumar…
Question Number 202530 by Calculusboy last updated on 28/Dec/23 Answered by MathematicalUser2357 last updated on 29/Dec/23 $$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{log}\left(\mathrm{1}+{x}\right)^{\mathrm{1}+{x}} −{x}}{{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\partial^{\mathrm{2}} }{\partial^{\mathrm{2}} {x}}\left(\mathrm{log}\left(\mathrm{1}+{x}\right)^{\mathrm{1}+{x}}…
Question Number 202514 by MrGHK last updated on 28/Dec/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 202462 by pticantor last updated on 27/Dec/23 Answered by witcher3 last updated on 27/Dec/23 $$\mathrm{p}^{\mathrm{n}} −\mathrm{1}=\left(\mathrm{p}−\mathrm{1}\right)\left(\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\mathrm{p}^{\mathrm{k}} \right) \\ $$$$\Rightarrow\mathrm{p}.\left(\mathrm{p}^{\mathrm{n}} \right)+\left(\mathrm{p}−\mathrm{1}\right).\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}}…
Question Number 202490 by Calculusboy last updated on 27/Dec/23 Commented by aleks041103 last updated on 27/Dec/23 $${is}\:\left\{{x}\right\}\:{the}\:{whole}\:{part}\:{or}\:{the}\:{fractional}\:{part}? \\ $$ Commented by Calculusboy last updated on…
Question Number 202459 by MATHEMATICSAM last updated on 27/Dec/23 $$\mathrm{If}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{of}\:\mathrm{two}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${x}^{\mathrm{2}} \:−\:{lx}\:+\:{m}\:=\:\mathrm{0}\:\mathrm{is}\:\mathrm{1}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$${l}^{\mathrm{2}} \:+\:\mathrm{4}{m}^{\mathrm{2}} \:=\:\left(\mathrm{1}\:+\:\mathrm{2}{m}\right)^{\mathrm{2}} \:. \\ $$ Answered by aleks041103 last updated…
Question Number 202485 by Calculusboy last updated on 27/Dec/23 Commented by MathematicalUser2357 last updated on 28/Dec/23 $$\mathrm{Can}'\mathrm{t}\:\mathrm{integrate}\:\mathrm{in}\:\mathrm{range}\:\left[−\infty,\:\infty\right] \\ $$ Commented by TheHoneyCat last updated on…
Question Number 202477 by MATHEMATICSAM last updated on 27/Dec/23 $$\mathrm{If}\:{x}\:=\:\frac{\sqrt{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }\:+\:\sqrt{{a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} }}{\:\sqrt{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }\:−\:\sqrt{{a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} }}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$${b}^{\mathrm{2}} {x}^{\mathrm{2}} \:−\:\mathrm{2}{a}^{\mathrm{2}} {x}\:+\:{b}^{\mathrm{2}} \:=\:\mathrm{0}. \\…
Question Number 202473 by hardmath last updated on 27/Dec/23 $$\mathrm{Find}: \\ $$$$\mathrm{1}.\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{2}^{\boldsymbol{\mathrm{n}}} \:\centerdot\:\mathrm{n}!}{\left(\mathrm{2n}\right)!}\:\:\:\:\:\:\:\:\:\:\mathrm{2}.\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left(\mathrm{x}\:+\:\mathrm{5}\right)^{\boldsymbol{\mathrm{n}}} }{\mathrm{3}^{\boldsymbol{\mathrm{n}}+\mathrm{1}} \:\centerdot\:\mathrm{n}\:\centerdot\:\mathrm{ln}\left(\mathrm{n}\right)} \\ $$ Terms of Service Privacy…