Question Number 111208 by Aziztisffola last updated on 02/Sep/20 $$\:\mathrm{let}\:{p}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}\:\mathrm{s}.\mathrm{t}\:{p}\geqslant\mathrm{7}\:\mathrm{and}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}=\underset{{p}−\mathrm{1}\:{times}} {\mathrm{333}……\mathrm{3}} \\ $$$$\:\mathrm{Show}\:\mathrm{that}\:\mathrm{11}\mid{a}. \\ $$ Commented by mr W last updated on 02/Sep/20…
Question Number 176750 by Tawa11 last updated on 26/Sep/22 $$\mathrm{Is}\:\mathrm{there}\:\mathrm{general}\:\mathrm{form}\:\mathrm{of} \\ $$$$\alpha^{\mathrm{n}} \:\:\:+\:\:\:\beta^{\mathrm{n}} \:\:\:\:\:\mathrm{and}\:\:\:\:\:\:\alpha^{\mathrm{n}} \:\:\:−\:\:\:\beta^{\mathrm{n}} \:\:\:\:\:???? \\ $$$$ \\ $$$$\mathrm{e}.\mathrm{g}:\:\:\:\:\alpha^{\mathrm{3}} \:\:\:+\:\:\:\beta^{\mathrm{3}} \:\:\:\:=\:\:\:\:\left(\alpha\:\:\:+\:\:\:\beta\right)\left[\left(\alpha\:\:\:+\:\:\:\beta\right)^{\mathrm{2}} \:\:\:−\:\:\:\mathrm{3}\alpha\beta\right] \\ $$…
Question Number 176746 by mr W last updated on 26/Sep/22 $${a}_{{n}+\mathrm{2}} −\mathrm{5}{a}_{{n}+\mathrm{1}} +\mathrm{6}{a}_{{n}} =\mathrm{3}{n}+\mathrm{5}^{{n}} \\ $$$${a}_{\mathrm{1}} =\mathrm{1},\:{a}_{\mathrm{2}} =\mathrm{0} \\ $$$${find}\:{a}_{{n}} \\ $$ Answered by FongXD…
Question Number 176727 by Shrinava last updated on 25/Sep/22 $$\mathrm{In}\:\:\:\bigtriangleup\mathrm{ABC}\:\:\:\mathrm{the}\:\mathrm{following}\:\mathrm{relationship} \\ $$$$\mathrm{holds}: \\ $$$$\mathrm{6r}\:\:\underset{\boldsymbol{\mathrm{cyc}}} {\sum}\:\frac{\mathrm{r}_{\boldsymbol{\mathrm{a}}} }{\mathrm{s}\:+\:\mathrm{n}_{\boldsymbol{\mathrm{a}}} }\:\:+\:\:\underset{\boldsymbol{\mathrm{cyc}}} {\sum}\:\mathrm{n}_{\boldsymbol{\mathrm{a}}} \:\geqslant\:\mathrm{3s} \\ $$ Terms of Service Privacy…
Question Number 176718 by Frix last updated on 25/Sep/22 $${x}^{\mathrm{3}} +{y}^{\mathrm{3}} ={z}^{\mathrm{2}} \\ $$$${x}^{\mathrm{3}} +{z}^{\mathrm{3}} ={y}^{\mathrm{2}} \\ $$$${y}^{\mathrm{3}} +{z}^{\mathrm{3}} ={x}^{\mathrm{2}} \\ $$$$\mathrm{obviously}\:{x}={y}={z}=\mathrm{0}\vee\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{trying}\:\mathrm{to}\:\mathrm{totally}\:\mathrm{solve}\:\mathrm{it} \\…
Question Number 111175 by 9696147350 last updated on 02/Sep/20 Answered by ajfour last updated on 02/Sep/20 $$\sqrt{{x}}={t} \\ $$$${t}^{\mathrm{4}} −\frac{\mathrm{3}{t}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0}\:\:\:\:,\:\:{let} \\ $$$$\left({t}^{\mathrm{2}} +{pt}+{q}\right)\left({t}^{\mathrm{2}} −{pt}+\frac{\mathrm{1}}{\mathrm{2}{q}}\right)=\mathrm{0} \\…
Question Number 176684 by peter frank last updated on 25/Sep/22 Commented by peter frank last updated on 25/Sep/22 $$\mathrm{thank}\:\mathrm{you} \\ $$ Commented by Rasheed.Sindhi last…
Question Number 176676 by Rasheed.Sindhi last updated on 24/Sep/22 $$\mathrm{If}\:\:\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }=\mathrm{1}, \\ $$$$\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{x}^{\mathrm{5}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{5}} }=−\left(\mathrm{x}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }\right) \\ $$ Answered by mr…
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