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Category: Algebra

Question-69644

Question Number 69644 by ahmadshahhimat775@gmail.com last updated on 26/Sep/19 Answered by MJS last updated on 26/Sep/19 $$\mathrm{let} \\ $$$${x}=\alpha \\ $$$${y}=\beta−\sqrt{\gamma} \\ $$$${z}=\beta+\sqrt{\gamma} \\ $$$$\begin{cases}{\alpha+\mathrm{2}\beta−\mathrm{12}=\mathrm{0}}\\{\alpha^{\mathrm{2}}…

Question-69645

Question Number 69645 by ahmadshahhimat775@gmail.com last updated on 26/Sep/19 Answered by Rasheed.Sindhi last updated on 26/Sep/19 $$\sqrt{{x}+\sqrt{{x}+\sqrt{{x}….}}}=\mathrm{2} \\ $$$$\left(\sqrt{{x}+\sqrt{{x}+\sqrt{{x}….}}}\right)=\left(\mathrm{2}\right)^{\mathrm{2}} \\ $$$${x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}…}}}=\mathrm{4} \\ $$$${x}+\mathrm{2}=\mathrm{4} \\ $$$${x}=\mathrm{2}…

x-2-9-3x-5-x-3-x-1-x-3-x-1-Find-solution-

Question Number 135172 by bemath last updated on 11/Mar/21 $$\left(\mathrm{x}^{\mathrm{2}} −\mathrm{9}\right)^{\mathrm{3x}+\mathrm{5}} \:=\:\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{x}−\mathrm{1}} .\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{x}−\mathrm{1}} \\ $$$$\mathrm{Find}\:\mathrm{solution} \\ $$ Answered by john_santu last updated on 11/Mar/21 $$\Rightarrow\left({x}^{\mathrm{2}}…

1-2-3-3-3-2-4-3-3-5-3-4-6-3-5-

Question Number 135170 by bramlexs22 last updated on 11/Mar/21 $$\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{4}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{5}}{\mathrm{3}^{\mathrm{4}} }+\frac{\mathrm{6}}{\mathrm{3}^{\mathrm{5}} }+…\:=?\: \\ $$$$ \\ $$ Answered by bemath last updated on 11/Mar/21…

Question-4075

Question Number 4075 by Yozzii last updated on 27/Dec/15 Answered by prakash jain last updated on 28/Dec/15 $$\:^{\mathrm{2}{r}} {C}_{{r}} =\frac{\mathrm{2}{r}!}{{r}!{r}!} \\ $$$$\mathrm{2}^{{r}} \centerdot{r}!=\mathrm{2}\centerdot\mathrm{4}\centerdot\mathrm{6}\centerdot…\centerdot\mathrm{2}{r} \\ $$$$\mathrm{2}{r}!=\mathrm{1}\centerdot\mathrm{2}\centerdot\mathrm{3}\centerdot\mathrm{4}\centerdot…\centerdot\left(\mathrm{2}{r}−\mathrm{1}\right)\centerdot\mathrm{2}{r}…

Question-69594

Question Number 69594 by ahmadshahhimat775@gmail.com last updated on 25/Sep/19 Answered by MJS last updated on 25/Sep/19 $$\mathrm{2}\leqslant{n}\leqslant\mathrm{4}:\:\mathrm{2}^{{n}!} <\mathrm{2}^{{n}} ! \\ $$$$\mathrm{5}\leqslant{n}:\:\mathrm{2}^{{n}!} >\mathrm{2}^{{n}} ! \\ $$$$\mathrm{ln}\:\mathrm{2}^{{n}!}…