Category: Algebra

Question Number 207226 by hardmath last updated on 10/May/24 $$\mathrm{Find}:\frac{{(2-\sqrt{2})}^8\cdot {(6+4\sqrt{2})}^8}{{(2+\sqrt{2})}^8}=?$$ Answered by Sutrisno last updated on 10/May/24 $${\left(\frac{(2-\sqrt{2})(6+4\sqrt{2})}{2+\sqrt{2}}\right)}^8$$$$\left(\frac{\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\left(\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}\right)}{\mathrm{2}+\sqrt{\mathrm{2}}}.\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{2}}}\right)^{\mathrm{8}}…

Question Number 207279 by hardmath last updated on 10/May/24 $$\mathrm{If}\mathrm{z}=\mathbf{i}-1$$$$\mathrm{Find}{\mathbf{z}}^{-100}=?$$ Commented by A5T last updated on 10/May/24 $$Q207220?$$…

Question Number 207274 by hardmath last updated on 10/May/24 $${\log }_{\mathbf{abc}}\mathrm{a}=2\mathrm{and}{\log }_{\mathbf{abc}}\mathrm{b}=3$$$$\mathrm{find}:{\log }_{\mathbf{abc}}\mathrm{c}=?$$ Answered by MM42 last updated on 10/May/24 $${log}_{{abc}}…

Question Number 207270 by hardmath last updated on 10/May/24 $$\{\begin{array}{l}\mid \mathrm{x}\mid +\mathrm{y}-1=0\\ \mathrm{x}-\mathrm{y}-1=0\end{array}\mathrm{find}:2\mathrm{x}-3\mathrm{y}=?$$ Answered by A5T last updated on 10/May/24 $$y=1-\mid x\mid =x-1$$$$\Rightarrow{x}+\mid{x}\mid=\mathrm{2}\Rightarrow\sqrt{{x}^{\mathrm{2}} }=\mathrm{2}−{x}\Rightarrow{x}^{\mathrm{2}} =\mathrm{4}+{x}^{\mathrm{2}} −\mathrm{4}{x}…

Question Number 207262 by hardmath last updated on 10/May/24 $${\mathrm{a}}_{\mathbf{n}}-\mathrm{number}\mathrm{series}$$$${\mathrm{a}}_1=5$$$$\mathrm{d}=3$$$$\mathrm{a}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{a}_{\mathrm{1}} ^{\mathrm{2}} \:+\:\mathrm{a}_{\mathrm{4}} ^{\mathrm{2}} −\mathrm{a}_{\mathrm{3}} ^{\mathrm{2}}…