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Question-171296




Question Number 171296 by dragan91 last updated on 11/Jun/22
Commented by Rasheed.Sindhi last updated on 13/Jun/22
For x>5    xy^3 +xy^3 +2022<20x!+22y!  For y>5    xy^3 +xy^3 +2022<20x!+22y!  this means we′ve to search values  between 0 & 5 for x,y.   ....  x=5,y=3 Ans
$${For}\:{x}>\mathrm{5}\:\:\:\:{xy}^{\mathrm{3}} +{xy}^{\mathrm{3}} +\mathrm{2022}<\mathrm{20}{x}!+\mathrm{22}{y}! \\ $$$${For}\:{y}>\mathrm{5}\:\:\:\:{xy}^{\mathrm{3}} +{xy}^{\mathrm{3}} +\mathrm{2022}<\mathrm{20}{x}!+\mathrm{22}{y}! \\ $$$${this}\:{means}\:{we}'{ve}\:{to}\:{search}\:{values} \\ $$$${between}\:\mathrm{0}\:\&\:\mathrm{5}\:{for}\:{x},{y}.\: \\ $$$$…. \\ $$$${x}=\mathrm{5},{y}=\mathrm{3}\:{Ans} \\ $$

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