Category: Algebra

Question Number 194295 by Abdullahrussell last updated on 02/Jul/23 Answered by BaliramKumar last updated on 02/Jul/23 $$1!\times 2!\times 3!\times \dots \dots \dots \dots \dots \dots \times 2023!$$$$1^{2023}\times 2^{2022}\times 3^{2021}\times \dots \dots \dots \dots \dots \dots \times {2023}^1$$$$\mathrm{1}^{\mathrm{2023}} ×…×\mathrm{5}^{\mathrm{2019}}…

Question Number 194170 by Rupesh123 last updated on 29/Jun/23 Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 194163 by York12 last updated on 29/Jun/23 $$\mathit{L}\mathit{e}\mathit{t}\mathit{a},\mathit{b},\mathit{c}\mathit{b}\mathit{e}\mathit{p}\mathit{o}\mathit{s}\mathit{i}\mathit{t}\mathit{i}\mathit{v}\mathit{e}\mathit{r}\mathit{e}\mathit{a}\mathit{l}\mathit{n}\mathit{u}\mathit{m}\mathit{b}\mathit{e}\mathit{r}\mathit{s}$$$$\mathit{p}\mathit{r}\mathit{o}\mathit{v}\mathit{e}\mathit{t}\mathit{h}\mathit{a}\mathit{t}$$$$\frac{\mathit{a}}{\mathit{b}}+\frac{\mathit{b}}{\mathit{c}}+\frac{\mathit{c}}{\mathit{a}}+\frac{3^3\sqrt{abc}}{\mathit{a}+\mathit{b}+\mathit{c}}⩾4$$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 194140 by MM42 last updated on 28/Jun/23 $$proveit:$$$${\left(\begin{array}{c}n\\ k\end{array}\right)}^2⩾\left(\begin{array}{c}n\\ k-1\end{array}\right)\times \left(\begin{array}{c}n\\ k+1\end{array}\right);1⩽k⩽n-1$$$$$$ Answered by witcher3 last updated on 28/Jun/23 $$\left(\frac{\mathrm{n}!}{\mathrm{k}!.\left(\mathrm{n}−\mathrm{k}\right)!}\right)^{\mathrm{2}}…

Question Number 194113 by Rupesh123 last updated on 28/Jun/23 Answered by Subhi last updated on 28/Jun/23 $$$$$$\frac{a(a-1)(a-2)\dots ..}{a(a-2)(a-4)\dots ..}.\frac{b(b-2)(b-4)\dots }{b(b-1)(b-2)\dots .}$$$$=\frac{(a-1)(a-3)(a-5)\dots ..}{(b-1)(b-3)(b-5)\dots ..}$$$$24canbethemultiplicationofn,n+2$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{4}\right),\left(\mathrm{6}\right)…