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Question-96231

Question Number 96231 by joki last updated on 30/May/20 Commented by prakash jain last updated on 30/May/20 $${x}^{\mathrm{3}} {y}−\mathrm{3}{xy}^{\mathrm{3}} =\mathrm{2}{a}^{\mathrm{4}} \\ $$$$\frac{{d}}{{dx}}\left({x}^{\mathrm{3}} {y}−\mathrm{3}{xy}^{\mathrm{3}} \right)=\frac{{d}}{{dx}}\mathrm{2}{a}^{\mathrm{4}} \\…

prove-that-n-0-1-n-n-1-ln2-

Question Number 161755 by mkam last updated on 22/Dec/21 $${prove}\:{that}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:=\:{ln}\mathrm{2} \\ $$ Answered by FelipeLz last updated on 22/Dec/21 $${f}\left({x}\right)\:=\:\mathrm{ln}\left({x}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:{f}'\left({x}\right)\:=\:\frac{\mathrm{1}}{{x}+\mathrm{1}}…

Question-161697

Question Number 161697 by mnjuly1970 last updated on 21/Dec/21 Answered by aleks041103 last updated on 21/Dec/21 $${let}\:{n}=\lfloor{x}\rfloor\:{and}\:{r}={x}−\lfloor{x}\rfloor\Rightarrow\mathrm{0}\leqslant{r}<\mathrm{1} \\ $$$$\Rightarrow\mathrm{0}\leqslant\sqrt{{r}}<\mathrm{1} \\ $$$$\sqrt{\mathrm{2}{n}+{r}}−\sqrt{{r}}=\mathrm{1} \\ $$$$\Rightarrow\sqrt{\mathrm{2}{n}+{r}}=\mathrm{1}+\sqrt{{r}} \\ $$$$\Rightarrow\mathrm{1}\leqslant\sqrt{\mathrm{2}{n}+{r}}<\mathrm{2}…

Question-96155

Question Number 96155 by pticantor last updated on 30/May/20 Commented by prakash jain last updated on 30/May/20 $${c}=\sqrt{\mathrm{4}−\sqrt{\mathrm{5}+{c}}} \\ $$$${c}^{\mathrm{2}} =\mathrm{4}−\sqrt{\mathrm{5}+{c}} \\ $$$${c}^{\mathrm{2}} −\mathrm{4}=−\sqrt{\mathrm{5}+{c}} \\…

xy-y-2-x-2-e-x-y-xe-x-y-2-x-y-xye-x-1-3x-y-3-y-3-3x-y-xye-x-0-y-y-2-3x-1-xe-x-0-y-3x-xe-x-1-

Question Number 96140 by Farruxjano last updated on 30/May/20 $$\boldsymbol{{xy}}'+\boldsymbol{{y}}^{\mathrm{2}} =\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{e}}^{\boldsymbol{{x}}} \:\Rightarrow\:\boldsymbol{{y}}'=\boldsymbol{{xe}}^{\boldsymbol{{x}}} −\frac{\boldsymbol{{y}}^{\mathrm{2}} }{\boldsymbol{{x}}}\:\Rightarrow \\ $$$$\boldsymbol{{y}}=\boldsymbol{{xye}}^{\boldsymbol{{x}}} −\frac{\mathrm{1}}{\mathrm{3}\boldsymbol{{x}}}\centerdot\boldsymbol{{y}}^{\mathrm{3}} \Rightarrow\:\frac{\boldsymbol{{y}}^{\mathrm{3}} }{\mathrm{3}\boldsymbol{{x}}}+\boldsymbol{{y}}−\boldsymbol{{xye}}^{\boldsymbol{{x}}} =\mathrm{0} \\ $$$$\boldsymbol{{y}}\left(\frac{\boldsymbol{{y}}^{\mathrm{2}} }{\mathrm{3}\boldsymbol{{x}}}+\mathrm{1}−\boldsymbol{{xe}}^{\boldsymbol{{x}}} \right)=\mathrm{0}\:\Rightarrow\:\boldsymbol{{y}}=\pm\sqrt{\mathrm{3}\boldsymbol{{x}}\left(\boldsymbol{{xe}}^{\boldsymbol{{x}}}…