Question Number 144537 by loveineq last updated on 26/Jun/21 $$\mathrm{Let}\:{a},{b}>\mathrm{0}\:\mathrm{and}\:{a}+{b}\:=\:\mathrm{2}.\:\mathrm{Prove}\:\mathrm{that}\:\:\:\:\:\:\:\:\:\: \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} }{\mathrm{2}}−\mathrm{2}\left(\mathrm{1}−{ab}\right)\:\geqslant\:\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:\:\:\:\:\:\:\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}\left(\mathrm{1}−{ab}\right)\:\leqslant\:\mathrm{1} \\ $$ Answered by mnjuly1970 last updated…
Question Number 13438 by Nayon last updated on 20/May/17 $$\mathrm{7}{x}^{\mathrm{5}} −\mathrm{4}{x}^{\mathrm{4}} +\mathrm{9}{x}^{\mathrm{3}} +\mathrm{12}{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{9}=\mathrm{0} \\ $$$${How}\:{many}\:{roots}\:{of}\:{this}\:{equation} \\ $$$${are}\:{Negative}? \\ $$$$ \\ $$ Answered by mrW1…
Question Number 78948 by TawaTawa last updated on 21/Jan/20 $$\mathrm{Prove}\:\mathrm{by}\:\mathrm{mathematical}\:\mathrm{induction}\:\mathrm{that}. \\ $$$$\:\:\:\mathrm{n}^{\mathrm{4}} \:+\:\mathrm{4n}^{\mathrm{2}} \:+\:\mathrm{11}\:\:\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{16} \\ $$ Commented by john santu last updated on 21/Jan/20 $${let}\:{p}\left({n}\right)\:=\:{n}^{\mathrm{4}}…
Question Number 13412 by tawa tawa last updated on 19/May/17 $$\mathrm{e}^{−\mathrm{kN}} \:−\:\mathrm{kN}\:\:−\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{N} \\ $$ Answered by mrW1 last updated on 20/May/17 $${e}^{−{kN}} ={kN}+\mathrm{1}…
Question Number 144468 by mathdanisur last updated on 25/Jun/21 Commented by Rasheed.Sindhi last updated on 26/Jun/21 $${It}'{s}\:{sufficient}\:{to}\:{say}\:'{natural}\:{numbers}' \\ $$$${instead}\:{of}\:'{positive}\:{natural}\:{numbers}'. \\ $$ Terms of Service Privacy…
Question Number 144470 by mathdanisur last updated on 25/Jun/21 $${Solve}\:{for}\:{real}\:{positive}\:{numbers}\:{the} \\ $$$${equation}: \\ $$$$\boldsymbol{{z}}^{\boldsymbol{{log}}\left(\mathrm{3}\right)} \:+\:\boldsymbol{{z}}^{\boldsymbol{{log}}\left(\mathrm{4}\right)} \:+\:\boldsymbol{{z}}^{\boldsymbol{{log}}\left(\mathrm{5}\right)} \:=\:\boldsymbol{{z}}^{\boldsymbol{{log}}\left(\mathrm{6}\right)} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 78931 by mr W last updated on 21/Jan/20 $${if}\:{x}+\frac{\mathrm{1}}{{x}}={a} \\ $$$${find}\:{x}^{{n}} +\frac{\mathrm{1}}{{x}^{{n}} }=? \\ $$ Commented by mathmax by abdo last updated on…
Question Number 13395 by Tinkutara last updated on 19/May/17 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{integers}\:{n}\:\mathrm{for}\:\mathrm{which} \\ $$$${n}^{\mathrm{2}} \:+\:\mathrm{96}\:\mathrm{is}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{square}. \\ $$ Answered by ajfour last updated on 19/May/17 $$\left({n}+{m}\right)^{\mathrm{2}} ={n}^{\mathrm{2}} +\mathrm{96}…
Question Number 13391 by Tinkutara last updated on 19/May/17 $$\mathrm{A}\:\mathrm{four}\:\mathrm{digit}\:\mathrm{number}\:\mathrm{has}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{properties}: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{It}\:\mathrm{is}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{square} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{The}\:\mathrm{first}\:\mathrm{two}\:\mathrm{digits}\:\mathrm{are}\:\mathrm{equal} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{The}\:\mathrm{last}\:\mathrm{two}\:\mathrm{digits}\:\mathrm{are}\:\mathrm{equal}. \\ $$$$\mathrm{Find}\:\mathrm{all}\:\mathrm{such}\:\mathrm{numbers}. \\ $$ Answered by RasheedSindhi…
Question Number 13394 by Tinkutara last updated on 19/May/17 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{any}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{centre}\:\left(\sqrt{\mathrm{2}},\:\sqrt{\mathrm{3}}\right) \\ $$$$\mathrm{cannot}\:\mathrm{pass}\:\mathrm{through}\:\mathrm{more}\:\mathrm{than}\:\mathrm{one} \\ $$$$\mathrm{lattice}\:\mathrm{point}.\:\left[\mathrm{Lattice}\:\mathrm{points}\:\mathrm{are}\:\mathrm{points}\right. \\ $$$$\mathrm{in}\:\mathrm{cartesian}\:\mathrm{plane},\:\mathrm{whose}\:\mathrm{abscissa}\:\mathrm{and} \\ $$$$\left.\mathrm{ordinate}\:\mathrm{both}\:\mathrm{are}\:\mathrm{integers}.\right] \\ $$ Answered by mrW1 last updated…