Question Number 157184 by cortano last updated on 20/Oct/21 $${p}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{3}{ax}^{\mathrm{2}} +\left(\mathrm{3}{a}^{\mathrm{2}} +\mathrm{1}\right){x}−\left({a}^{\mathrm{3}} +{a}\right) \\ $$$${p}\left(\mathrm{2}\right)<\mathrm{0} \\ $$$${prove}\:{that}\:\mathrm{2}<{a}<\mathrm{3} \\ $$ Answered by ghimisi last updated…
Question Number 26115 by ktomboy1992 last updated on 21/Dec/17 $$\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=\mathrm{3}\:\mathrm{thrn}\:\mathrm{find}\:\mathrm{the}\:\mathrm{valu}\:\mathrm{of}\left(\:\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} \\ $$ Answered by $@ty@m last updated on 20/Dec/17 $$\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{2}=\mathrm{3}−\mathrm{2}=\mathrm{1}…
Question Number 91647 by mr W last updated on 02/May/20 $${solve} \\ $$$${x}\lfloor{x}\lfloor{x}\lfloor{x}\rfloor\rfloor\rfloor=\mathrm{2020} \\ $$ Answered by frc2crc last updated on 03/May/20 $${i}\:{used}\:{desmos}\:{x}\approx−\mathrm{6}.\mathrm{623} \\ $$…
Question Number 157163 by MathSh last updated on 20/Oct/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{solution} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:=\:\mathrm{7}\centerdot\left(\mathrm{14xy}\:+\:\mathrm{3}\right) \\ $$ Commented by Rasheed.Sindhi last updated on 22/Oct/21…
Question Number 26067 by Joel578 last updated on 19/Dec/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\frac{\mathrm{2}\:+\:\mathrm{3}^{\mathrm{2}} }{\mathrm{1}!\:+\:\mathrm{2}!\:+\:\mathrm{3}!\:+\:\mathrm{4}!}\:+\:\frac{\mathrm{3}\:+\:\mathrm{4}^{\mathrm{2}} }{\mathrm{2}!\:+\:\mathrm{3}!\:+\:\mathrm{4}!\:+\:\mathrm{5}!}\:+\:…\:+\:\frac{\mathrm{2013}\:+\:\mathrm{2014}^{\mathrm{2}} }{\mathrm{2012}!\:+\:\mathrm{2013}!\:+\:\mathrm{2014}!\:+\:\mathrm{2015}!} \\ $$ Answered by iv@0uja last updated on 19/Dec/17 $$\:\:\:\:\:\frac{\mathrm{2}\:+\:\mathrm{3}^{\mathrm{2}}…
Question Number 91599 by john santu last updated on 01/May/20 Commented by mr W last updated on 01/May/20 $${x}=\mathrm{0}\:{is}\:{a}\:{solution}. \\ $$$${since}\:\frac{{d}\left({LHS}\right)}{{dx}}>\frac{{d}\left({RHS}\right)}{{dx}}=\mathrm{4}, \\ $$$${x}=\mathrm{0}\:{is}\:{also}\:{the}\:{only}\:{one}\:{solution}. \\ $$…
Question Number 157120 by cortano last updated on 20/Oct/21 $$\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\left(\mathrm{3}{n}+\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{2}\right)}\right)=? \\ $$ Answered by puissant last updated on 20/Oct/21 $$\Omega=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\left(\mathrm{3}{n}+\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{2}\right)}\right)=\frac{\mathrm{1}}{\mathrm{9}}\underset{{n}=\mathrm{0}} {\overset{\infty}…
Question Number 26053 by tawa tawa last updated on 18/Dec/17 $$\mathrm{If}\:\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{9x}\:+\:\mathrm{2}\:=\:\mathrm{0}\:\:\mathrm{and}\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{kx}\:+\:\mathrm{5}\:=\:\mathrm{0}\:\:\mathrm{have}\:\mathrm{a}\:\mathrm{common}\:\mathrm{root},\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\mathrm{2k}^{\mathrm{2}} \:+\:\mathrm{63k}\:−\:\mathrm{414}\:=\:\mathrm{0}\:,\:\:\mathrm{hence}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{k}\:\:\mathrm{such}\:\mathrm{that}\:\mathrm{k}\:>\:\mathrm{9}.\mathrm{3} \\ $$ Answered by kaivan.ahmadi last updated on 18/Dec/17…
Question Number 26042 by sanjib bibhar last updated on 18/Dec/17 $${ax}^{\mathrm{2}} −{bx}=\mathrm{0} \\ $$ Commented by sanjib bibhar last updated on 19/Dec/17 $${thanks} \\ $$…
Question Number 91560 by ar247 last updated on 01/May/20 $${x}=\frac{\mathrm{1}+\sqrt{\mathrm{2004}}}{\mathrm{2}} \\ $$$$\mathrm{4}{x}^{\mathrm{3}} −\mathrm{2007}{x}−\mathrm{2000}=? \\ $$ Commented by Prithwish Sen 1 last updated on 01/May/20 $$\mathrm{x}^{\mathrm{3}}…