Category: Algebra

Question Number 98570 by HamraboyevFarruxjon last updated on 14/Jun/20 $$\mathit{a},\mathit{b},\mathit{c}>0\mathit{p}\mathit{r}\mathit{o}\mathit{v}\mathit{e}:$$$$\frac{\mathit{a}}{\sqrt{{\mathit{a}}^2+8\mathit{b}\mathit{c}}}+\frac{\mathit{b}}{\sqrt{{\mathit{b}}^2+8\mathit{a}\mathit{c}}}+\frac{\mathit{c}}{\sqrt{{\mathit{c}}^2+8\mathit{a}\mathit{b}}}⩾1$$$$\mathit{h}\mathit{e}\mathit{l}\mathit{p}\mathit{p}\mathit{l}\mathit{e}\mathit{a}\mathit{s}\mathit{e}\dots$$ Commented by MJS last updated on…

Question Number 164077 by mathlove last updated on 13/Jan/22 Answered by cortano1 last updated on 13/Jan/22 $$\left(1\right)x+y+xy+1=4\Rightarrow (x+1)(y+1)=4$$$$\left(2\right)y+z+yz+1=6\Rightarrow (y+1)(z+1)=6$$$$\left(3\right)z+x+zx+1=8\Rightarrow (z+1)(x+1)=8$$$$\left(1\right)\times \left(2\right)\times \left(3\right)\Rightarrow (x+1)(y+1)(z+1)=\pm \sqrt{4\times 4\times 4\times 3}=\pm 8\sqrt{3}$$$$\:{z}+\mathrm{1}=\pm\mathrm{2}\sqrt{\mathrm{3}}…

Question Number 164071 by kdaramaths last updated on 13/Jan/22 Answered by Ar Brandon last updated on 13/Jan/22 $$Z=\underset{k=0}{\sum ^n}{e}^{ikx}=\frac{1-e^{i(n+1)x}}{1-e^{ix}}$$$$\:\:\:=\frac{{e}^{{i}\left(\frac{\left({n}+\mathrm{1}\right){x}}{\mathrm{2}}\right)}…

Question Number 164050 by mathlove last updated on 13/Jan/22 Answered by Ar Brandon last updated on 13/Jan/22 $$x-\frac{1}{8}>0\ne 1\Rightarrow x>\frac{1}{8}\ne \frac{9}{8}\dots .\left(1\right)$$$$x^2-\frac{1}{9}>0\Rightarrow x<-\frac{1}{3}\cup x>\frac{1}{3}\dots \left(2\right)$$$$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{2}−{x}\right)>\mathrm{0}\:\Rightarrow\mathrm{2}−{x}>\mathrm{0}\Rightarrow{x}<\mathrm{2}…\left(\mathrm{3}\right) \…

Question Number 164039 by ArielVyny last updated on 13/Jan/22 $$considerffunctionDf=[0,1]$$$$f\left(0\right)=f\left(1\right)c\in [0,\frac{1}{2}]showthatf\left(c\right)=f(c+\frac{1}{2})$$ Answered by Ar Brandon last updated on 13/Jan/22 $$f\left(0\right)=f\left(1\right)$$$${f}\left({c}+\mathrm{0}\right)={f}\left({c}+\mathrm{1}\right)\:,\:\mathrm{since}\:{f}\:<\:\mathrm{1}-\mathrm{periodic}…