Category: Algebra

Question Number 64404 by aliesam last updated on 17/Jul/19 Commented by Prithwish sen last updated on 17/Jul/19 $$\mathrm{x}+1=\sqrt{{(\mathrm{x}+1)}^2}=\sqrt{1+\mathrm{x}(\mathrm{x}+2)}=\sqrt{1+\mathrm{x}\sqrt{{(\mathrm{x}+2)}^2}}$$$$=\sqrt{1+\mathrm{x}\sqrt{1+(\mathrm{x}+1)(\mathrm{x}+3)}}=\sqrt{1+\mathrm{x}\sqrt{1+(\mathrm{x}+1)\sqrt{{(\mathrm{x}+3)}^2}}}$$$$=\:\sqrt{\mathrm{1}+\mathrm{x}\sqrt{\mathrm{1}+\left(\mathrm{x}+\mathrm{1}\right)\sqrt{\mathrm{1}+\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{x}+\mathrm{4}\right)}}\:}…

Question Number 129841 by Adel last updated on 20/Jan/21 $$\left\{{}_{\sqrt[3]{2^{2\mathrm{x}}}+\sqrt[3]{2^{4\mathrm{x}}}+4^{\mathrm{x}}=5}^{\sqrt[3]{2^{\mathrm{x}}}-2^{\mathrm{x}}=2}\right\}\Rightarrow 2^{\mathrm{x}}-8^{\mathrm{x}}=?$$$$\mathrm{pleas}\mathrm{solve}\mathrm{thes}$$ Commented by…

Question Number 64284 by Tawa1 last updated on 16/Jul/19 $$\mathrm{Find}\mathrm{all}\mathrm{the}\mathrm{integer}\mathrm{solution}{3\mathrm{x}}^2+1={4\mathrm{y}}^3$$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 64186 by Hope last updated on 15/Jul/19 Answered by Hope last updated on 15/Jul/19 $$\frac{{\left(\sqrt{2}\right)}^2}{x+y}+\frac{{\left(\sqrt{2}\right)}^2}{y+z}+\frac{{\left(\sqrt{2}\right)}^2}{z+x}⩾\frac{{(\sqrt{2}+\sqrt{2}+\sqrt{2})}^2}{2(x+y+z)}$$$$do⩾\frac{9\times 2}{2(x+y+z)}$$$$\frac{\mathrm{2}}{{x}+{y}}+\frac{\mathrm{2}}{{y}+{z}}+\frac{\mathrm{2}}{{z}+{x}}\geqslant\frac{\mathrm{9}}{{x}+{y}+{z}}\:{proved}…

Question Number 64174 by gajrajgchouhan last updated on 15/Jul/19 $$2^n/5^{2^n}+1infiniteseriessumffom0to$$$$infinity$$ Terms of Service Privacy Policy Contact: info@tinkutara.com