Category: Arithmetic

Question Number 185805 by Rupesh123 last updated on 27/Jan/23 Answered by MJS_new last updated on 28/Jan/23 $${(2+\sqrt{3})}^{2023}=x$$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}×\frac{{x}^{\mathrm{2}} +{x}^{−\mathrm{2}} +\mathrm{1}}{{x}+{x}^{+\mathrm{1}} }=\frac{\mathrm{1}+{n}^{\mathrm{2}} }{{n}}\:\Rightarrow\:{n}=\left(\frac{\sqrt{\mathrm{3}}{x}}{{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\pm\mathrm{1}}…

Question Number 120241 by olalekanoriyomi last updated on 30/Oct/20 $$\mathbf{C}orrection\mathit{t}\mathit{o}\mathit{t}\mathit{h}\mathit{e}\mathit{l}\mathit{a}\mathit{s}\mathit{t}\mathit{a}\mathit{s}\mathit{s}\mathit{i}\mathit{g}\mathit{n}\mathit{m}\mathit{e}\mathit{n}\mathit{t}$$$$$$$$\left(1\right)\frac{1}{{81}^{\mathit{x}-2}}={27}^{1-\mathit{x}}$$$$\frac{1}{3^{4(\mathit{x}-2)}}=3^{3(1-\mathit{x})}$$$$\frac{1}{3^{4\mathit{x}-8}}=3^{3-3\mathit{x}}$$$$\mathrm{3}^{−\mathrm{4}\boldsymbol{{x}}+\mathrm{8}} =\mathrm{3}^{\mathrm{3}−\mathrm{3}\boldsymbol{{x}}} \…

Question Number 185743 by Rupesh123 last updated on 26/Jan/23 Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 54322 by peter frank last updated on 02/Feb/19 Answered by kaivan.ahmadi last updated on 02/Feb/19 $$\mathrm{ax}+\mathrm{by}=\mathrm{c}\mathrm{and}\mathrm{bx}-\mathrm{ay}=\mathrm{d}\mathrm{are}\mathrm{perpendicular}$$$$\mathrm{abx}^{\mathrm{2}} +\left(\mathrm{b}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} \right)\mathrm{xy}−\mathrm{aby}^{\mathrm{2}} =\mathrm{cd} \…

Question Number 185269 by Rupesh123 last updated on 19/Jan/23 Answered by aleks041103 last updated on 19/Jan/23 $$p=\sqrt{n-8}+\sqrt{n+8}\Rightarrow n⩾8$$$$p^2=2n-2\sqrt{n^2-64}$$$$\Rightarrow\sqrt{\mathrm{4}\left({n}^{\mathrm{2}} −\mathrm{64}\right)}=\mathrm{2}{n}−{p}^{\mathrm{2}} \…