Question Number 202762 by Mastermind last updated on 02/Jan/24 Answered by shunmisaki007 last updated on 03/Jan/24 $$\boldsymbol{{A}}=\begin{bmatrix}{\mathrm{3}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{3}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{3}}\end{bmatrix} \\ $$$$\mathrm{adj}\left(\boldsymbol{{A}}\right)=\begin{bmatrix}{\begin{vmatrix}{\mathrm{3}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{3}}\end{vmatrix}}&{\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{3}}\end{vmatrix}}&{\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{3}}&{\mathrm{0}}\end{vmatrix}}\\{\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{3}}\end{vmatrix}}&{\begin{vmatrix}{\mathrm{3}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{3}}\end{vmatrix}}&{\begin{vmatrix}{\mathrm{3}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}\end{vmatrix}}\\{\begin{vmatrix}{\mathrm{0}}&{\mathrm{3}}\\{\mathrm{0}}&{\mathrm{0}}\end{vmatrix}}&{\begin{vmatrix}{\mathrm{3}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}\end{vmatrix}}&{\begin{vmatrix}{\mathrm{3}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{3}}\end{vmatrix}}\end{bmatrix}=\begin{bmatrix}{\mathrm{9}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{9}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{9}}\end{bmatrix} \\ $$$$\mathrm{adj}\left(\mathrm{adj}\left(\boldsymbol{{A}}\right)\right)=\begin{bmatrix}{\begin{vmatrix}{\mathrm{9}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{9}}\end{vmatrix}}&{\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{9}}\end{vmatrix}}&{\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{9}}&{\mathrm{0}}\end{vmatrix}}\\{\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{9}}\end{vmatrix}}&{\begin{vmatrix}{\mathrm{9}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{9}}\end{vmatrix}}&{\begin{vmatrix}{\mathrm{9}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}\end{vmatrix}}\\{\begin{vmatrix}{\mathrm{0}}&{\mathrm{9}}\\{\mathrm{0}}&{\mathrm{0}}\end{vmatrix}}&{\begin{vmatrix}{\mathrm{9}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}\end{vmatrix}}&{\begin{vmatrix}{\mathrm{9}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{9}}\end{vmatrix}}\end{bmatrix}=\begin{bmatrix}{\mathrm{27}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{27}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{27}}\end{bmatrix} \\ $$$$\mathrm{det}\left(\mathrm{adj}\left(\mathrm{adj}\left(\boldsymbol{{A}}\right)\right)\right)=\begin{vmatrix}{\mathrm{27}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{27}}&{\mathrm{0}}\\{\mathrm{0}}&{\:\mathrm{0}}&{\mathrm{27}}\end{vmatrix}=\mathrm{19},\mathrm{683} \\ $$$$\frac{\mathrm{det}\left(\mathrm{adj}\left(\mathrm{adj}\left(\boldsymbol{{A}}\right)\right)\right)}{\mathrm{5}}=\frac{\mathrm{19},\mathrm{683}}{\mathrm{5}}=\mathrm{3},\mathrm{936}\frac{\mathrm{3}}{\mathrm{5}}…
Question Number 202721 by Rasheed.Sindhi last updated on 01/Jan/24 $$\:\:\:\begin{array}{|c|}{\underset{\mathcal{HAPPY}\:\mathcal{NEW}\:\mathcal{YEAR}!} {\mathrm{2}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} +\centerdot\centerdot\centerdot\centerdot\centerdot+\mathrm{22}^{\mathrm{2}} =\mathrm{2024}}}\\\hline\end{array} \\ $$$$\begin{array}{|c|}{\:\:\:\mathrm{May}\:\mathrm{2024}\:\mathrm{be}\:\begin{cases}{\mathrm{war}-\mathrm{free}!}\\{\&}\\{{peaceful}!}\end{cases}\:\:\:}\\\hline\end{array} \\ $$$$ \\ $$ Terms of Service Privacy…
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Question Number 202082 by Abduljalal last updated on 19/Dec/23 Answered by AST last updated on 20/Dec/23 $${p}^{\mathrm{2}} −{p}−\mathrm{2}=\mathrm{0}\left({p}={x}^{\mathrm{3}} \right)\Rightarrow{p}=\mathrm{2}\:{or}\:−\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{3}} =\mathrm{2}\:{or}\:{x}^{\mathrm{3}} =−\mathrm{1}\Rightarrow{x}=\sqrt[{\mathrm{3}}]{\mathrm{2}};\sqrt[{\mathrm{3}}]{\mathrm{2}}{e}^{\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} ,\sqrt[{\mathrm{3}}]{\mathrm{2}}{e}^{\frac{\mathrm{4}\pi{i}}{\mathrm{3}}} \\…
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Question Number 201814 by Euclid last updated on 13/Dec/23 Answered by sajitha last updated on 13/Dec/23 $$\mathrm{4}^{{x}^{\mathrm{3}} } =\mathrm{4} \\ $$$${x}^{\mathrm{3}} =\mathrm{1} \\ $$$${x}^{\mathrm{3}} −\mathrm{1}=\mathrm{0}…
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Question Number 201751 by esmaeil last updated on 11/Dec/23 $${hi} \\ $$$${Is}\:\:{it}\:\:{possible}\:{drawing}\:{with}\:{naming} \\ $$$${by}\:{math}\:{editor}? \\ $$$$\left({sorry}\:{i}'{m}\:{not}\:{good}\:{in}\:{english}\right) \\ $$ Commented by Rasheed.Sindhi last updated on 11/Dec/23…
Question Number 201491 by tri26112004 last updated on 07/Dec/23 $$\mathrm{1}.\:{x}^{\mathrm{2}} −\mathrm{1}+\sqrt[{\mathrm{3}}]{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} }=\mathrm{10}{x} \\ $$$$\mathrm{2}.\:\sqrt{{x}−\frac{\mathrm{1}}{{x}}}+\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}={x} \\ $$$$\mathrm{3}.\:\sqrt{\mathrm{2}{x}−\frac{\mathrm{8}}{{x}}}+\mathrm{2}\sqrt{\mathrm{1}−\frac{\mathrm{2}}{{x}}}\geqslant{x} \\ $$$$\mathrm{4}.\:\sqrt{{x}^{\mathrm{2}} +{x}}+\sqrt{{x}+\mathrm{2}}\geqslant\sqrt{\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)} \\ $$$$\mathrm{5}.\:\left(\sqrt{{x}+\mathrm{5}}−\sqrt{{x}−\mathrm{3}}\right)\left(\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{15}}\right)\geqslant\mathrm{8} \\…