Category: Algebra

Question Number 27761 by shiv15031973@gmail.com last updated on 14/Jan/18 $$4{(2\mathrm{a}+\mathrm{b})}^2-{(\mathrm{a}-\mathrm{b})}^2$$ Answered by Rasheed.Sindhi last updated on 14/Jan/18 $$4{(2\mathrm{a}+\mathrm{b})}^2-{(\mathrm{a}-\mathrm{b})}^2$$$$=\mathrm{4}\left(\mathrm{4a}^{\mathrm{2}} +\mathrm{4ab}+\mathrm{b}^{\mathrm{2}}…

Question Number 158816 by HongKing last updated on 09/Nov/21 $$\mathrm{Prove}\mathrm{that}{2017}^{2017}\mathrm{and}{2017}^{2018}$$$$\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{two}$$$$\mathrm{perfect}\mathrm{squares}.$$$$$$ Terms of Service Privacy Policy Contact:…

Question Number 158813 by HongKing last updated on 09/Nov/21 $$\mathrm{Find}\mathrm{all}\mathrm{value}\mathit{\beta }>0\mathrm{such}\mathrm{that}:$$$$\underset{0}{\stackrel{+\mathrm{\infty }}{\int }}\frac{\mathrm{dx}}{{\mathrm{x}}^{2021\mathit{\beta }}+\ln \cdot (1+\beta \mathrm{x})}\mathrm{dx}<+\mathrm{\infty }$$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 158781 by HongKing last updated on 08/Nov/21 Answered by MJS_new last updated on 09/Nov/21 $$\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{\left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}{x}\right)\mathrm{cos}\:\mathrm{4}{x}}{\mathrm{1}+{a}\mathrm{sin}^{\mathrm{2}} \:{x}}{dx}=\mathrm{2}\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{\left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}{x}\right)\mathrm{cos}\:\mathrm{4}{x}}{\mathrm{1}+{a}\mathrm{sin}^{\mathrm{2}} \:{x}}…

Question Number 158774 by HongKing last updated on 08/Nov/21 Answered by MJS_new last updated on 08/Nov/21 $${\sin }^{4}x+{\cos }^{2}x={\sin }^{2}x+{\cos }^{4}x$$$$\Rightarrow \mathrm{we}\mathrm{have}$$$$\mathrm{2}\sqrt{\mathrm{7}+\mathrm{cos}\:\mathrm{4}{x}}+\sqrt{\mathrm{9}−\mathrm{cos}\:\mathrm{4}{x}}=\mathrm{6}\sqrt{\mathrm{2}}…