Question Number 180785 by Vynho last updated on 17/Nov/22 $${solve}\:\left(\mathrm{1}−{x}−{x}^{\mathrm{2}} …\right)\left(\mathrm{2}−{x}−{x}^{\mathrm{2}} …\right) \\ $$ Answered by Rasheed.Sindhi last updated on 17/Nov/22 $$\left(\:\mathrm{1}−\left({x}+{x}^{\mathrm{2}} +…\right)\:\right)\left(\:\mathrm{2}−\left({x}+{x}^{\mathrm{2}} +…\right)\:\right) \\…
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Question Number 180759 by depressiveshrek last updated on 16/Nov/22 $${Find}\:{all}\:{functions}\:{f}\::\:\mathbb{R}\rightarrow\mathbb{R}\:{for}\:{all}\:{x},\:{y}\:\in\mathbb{R}\:{such}\:{that} \\ $$$${f}\left({f}\left({x}−{y}\right)−{yf}\left({x}\right)\right)={xf}\left({y}\right) \\ $$ Answered by aleks041103 last updated on 17/Nov/22 $${x}=\mathrm{0} \\ $$$$\Rightarrow{f}\left({f}\left(−{y}\right)−{yf}\left(\mathrm{0}\right)\right)=\mathrm{0} \\…
Question Number 115218 by Ar Brandon last updated on 24/Sep/20 $$\mathrm{Show}\:\mathrm{that}\:\forall\mathrm{n}\in\mathbb{N},\:\forall\mathrm{u}_{\mathrm{0}} ,\mathrm{u}_{\mathrm{1}} ,…,\mathrm{u}_{\mathrm{n}} ,\mathrm{v}_{\mathrm{0}} ,\mathrm{v}_{\mathrm{1}} ,…\mathrm{v}_{\mathrm{n}} \in\mathbb{C} \\ $$$$\forall\mathrm{k}\leqslant\mathrm{n};\:\mathrm{u}_{\mathrm{k}} =\sum_{\mathrm{i}=\mathrm{0}} ^{\mathrm{k}} \begin{pmatrix}{\mathrm{k}}\\{\mathrm{i}}\end{pmatrix}\mathrm{v}_{\mathrm{i}} \Leftrightarrow\forall\mathrm{k}\leqslant\mathrm{n};\:\mathrm{v}_{\mathrm{k}} =\sum_{\mathrm{i}=\mathrm{0}} ^{\mathrm{k}}…
Question Number 180744 by harckinwunmy last updated on 16/Nov/22 Commented by cyathokoza last updated on 17/Nov/22 $$\boldsymbol{{I}}\:\boldsymbol{{TRY}}\:\:\:\boldsymbol{{THIS}}\:\boldsymbol{{THEN}}\:\boldsymbol{{SUBMIT}}.\:\boldsymbol{{THE}}\:\boldsymbol{{ANSWERS}} \\ $$ Commented by Acem last updated on…
Question Number 180728 by CrispyXYZ last updated on 16/Nov/22 $$\mathrm{Given}\:{x},\:{y},\:{z}\in\mathbb{R}^{+} \:\mathrm{such}\:\mathrm{that}\:{x}^{\mathrm{2}} −\mathrm{3}{xy}+\mathrm{4}{y}^{\mathrm{2}} −{z}=\mathrm{0}. \\ $$$$\mathrm{when}\:\frac{{xy}}{{z}}\:\mathrm{reaches}\:\mathrm{its}\:\mathrm{max}\:\mathrm{value},\:\mathrm{find}\:\mathrm{the}\:\mathrm{max}\:\mathrm{value}\:\mathrm{of} \\ $$$$\frac{\mathrm{2}}{{x}}+\frac{\mathrm{1}}{{y}}−\frac{\mathrm{2}}{{z}}. \\ $$ Answered by mr W last updated…
Question Number 49647 by maxmathsup by imad last updated on 08/Dec/18 $${let}\:{p}\left({x}\right)\:={x}^{\mathrm{2}{n}} \:−{x}^{{n}} \:+\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{determine}\:{the}\:{roots}\:{of}\:{p}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{factorize}\:{inside}\:{C}\left[{x}\right]\:{the}\:{polynom}\:{p}\left({x}\right)\:. \\ $$$$\left.\mathrm{3}\right){solve}\:{p}\left({x}\right)=\mathrm{0}\:\:{and}\:{p}\left({x}\right)\:=\mathrm{2} \\ $$ Commented by maxmathsup…
Question Number 49642 by maxmathsup by imad last updated on 08/Dec/18 $${if}\:\:{a}+{b}\:={s}\:{and}\:{a}^{\mathrm{3}} \:+{b}^{\mathrm{3}} \:={t}\:\:{find}\:{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \:\:{and}\:{a}^{\mathrm{4}} \:+{b}^{\mathrm{4}} \:{interms}\:{of}\:{s}\:{and}\:{t}\:. \\ $$ Answered by mr W last…
Question Number 49604 by Tawa1 last updated on 08/Dec/18 $$\mathrm{Show}\:\mathrm{that}:\:\:\:\:\:\:\:\:\:\:\:\frac{\left(\mathrm{a}\:+\:\mathrm{b}\right)^{\mathrm{2}} }{\mathrm{2}}\:\:\leqslant\:\:\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \\ $$ Answered by afachri last updated on 08/Dec/18 $$\mathrm{let}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({a}\:−\:{b}\right)^{\mathrm{2}} \:\:\geqslant\:\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{2}}…
Question Number 115135 by Algoritm last updated on 23/Sep/20 Commented by Dwaipayan Shikari last updated on 23/Sep/20 $$\frac{\mathrm{1}}{\mathrm{51}.\mathrm{52}…\mathrm{100}}=\frac{\mathrm{1}.\mathrm{2}.\mathrm{3}..}{\mathrm{100}!}=\frac{\mathrm{50}!}{\mathrm{100}!} \\ $$$$\mathrm{1}.\mathrm{3}.\mathrm{5}..\mathrm{9}…=\frac{\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}…}{\mathrm{2}^{\mathrm{100}} .\mathrm{50}!} \\ $$$$\frac{\mathrm{100}!}{\mathrm{2}^{\mathrm{50}} .\mathrm{50}!}.\frac{\mathrm{50}!}{\mathrm{100}!}=\mathrm{2}^{\mathrm{x}} \\…