Category: Arithmetic

Question Number 131010 by Boucatchou last updated on 31/Jan/21 $$Determineintegersa,bsuchthat\{\begin{array}{l}a+b=182\\ a\vee b=13\end{array}$$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 130817 by shaker last updated on 29/Jan/21 Answered by Olaf last updated on 29/Jan/21 $$\mathrm{argth}({\mathrm{sh}}^{4}x+{3\mathrm{sh}}^{2}x)=1$$$$\mathrm{sh}^{\mathrm{4}} {x}+\mathrm{3sh}^{\mathrm{2}} {x}\:=\:\mathrm{th}\left(\mathrm{1}\right)\:=\:\frac{{e}−\frac{\mathrm{1}}{{e}}}{{e}+\frac{\mathrm{1}}{{e}}}\:=\:\frac{{e}^{\mathrm{2}} −\mathrm{1}}{{e}^{\mathrm{2}} +\mathrm{1}}…

Question Number 65122 by aliesam last updated on 25/Jul/19 Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 64916 by Masumsiddiqui399@gmail.com last updated on 23/Jul/19 Commented by Tawa1 last updated on 23/Jul/19 $${(2+\sqrt{3})}^{\mathrm{x}}+{\left(\frac{1}{2+\sqrt{3}}\right)}^{\mathrm{x}}=4[\mathrm{since}2-\sqrt{3}=\frac{1}{2+\sqrt{3}}\mathrm{on}\mathrm{rationalise}]$$$${(2+\sqrt{3})}^{\mathrm{x}}+\frac{1}{{(2+\sqrt{3})}^{\mathrm{x}}}=4$$$$\mathrm{Let}\:\:\:\:\:\left(\mathrm{2}\:+\:\sqrt{\mathrm{3}}\right)^{\mathrm{x}}…

Question Number 64561 by aliesam last updated on 19/Jul/19 Commented by mathmax by abdo last updated on 20/Jul/19 $${let}\:{P}_{{n}} =\prod_{{k}=\mathrm{0}} ^{{n}} \left(\mathrm{1}+\sigma^{\mathrm{2}^{{k}} } \right)\:\Rightarrow{P}_{{n}} =\left(\mathrm{1}+\sigma\right)\left(\mathrm{1}+\sigma^{\mathrm{2}}…