Question Number 20550 by Tinkutara last updated on 28/Aug/17 $${Find}\:{the}\:{equation}\:{of}\:{circle}\:{in}\:{complex} \\ $$$${form}\:{which}\:{touches}\:{iz}\:+\:\bar {{z}}\:+\:\mathrm{1}\:+\:{i}\:=\:\mathrm{0} \\ $$$${and}\:{for}\:{which}\:{the}\:{lines}\:\left(\mathrm{1}\:−\:{i}\right){z}\:= \\ $$$$\left(\mathrm{1}\:+\:{i}\right)\bar {{z}}\:{and}\:\left(\mathrm{1}\:+\:{i}\right){z}\:+\:\left({i}\:−\:\mathrm{1}\right)\bar {{z}}\:−\:\mathrm{4}{i}\:=\:\mathrm{0} \\ $$$${are}\:{normals}. \\ $$ Answered by…
Question Number 151616 by mathdanisur last updated on 22/Aug/21 $$\mathrm{let}\:\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\boldsymbol{\lambda}+\mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\:\mathrm{and}\:\:\boldsymbol{\lambda}\geqslant\frac{-\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{solve}\:\mathrm{in}\:\mathbb{R}\:\:\:\mathrm{f}\left(\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\right)\:\leqslant\:\mathrm{0} \\ $$ Commented by mr W last updated on 22/Aug/21 $${then}\:{solution}\:{is}\:{x}\geqslant\lambda \\…
Question Number 86085 by Serlea last updated on 27/Mar/20 $$\mathrm{If}\:\mathrm{X}^{\mathrm{2}} +\mathrm{Y}^{\mathrm{2}} =\mathrm{10} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{XY}=\mathrm{5} \\ $$$$\mathrm{Find}\:\left(\mathrm{X}^{\mathrm{2}} −\mathrm{Y}^{\mathrm{2}} \right) \\ $$ Commented by jagoll last updated…
Question Number 151614 by mathdanisur last updated on 22/Aug/21 $$\boldsymbol{\Omega}\:=\underset{\:\mathrm{0}} {\overset{\:\mathrm{2}\boldsymbol{\pi}} {\int}}\frac{\mathrm{x}\:+\:\mathrm{tan}\left(\mathrm{sin}\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\lambda}\:+\:\mathrm{cos}\left(\boldsymbol{\mathrm{x}}\right)}\:\mathrm{dx}\:\:;\:\:\boldsymbol{\lambda}>\mathrm{1} \\ $$ Answered by ArielVyny last updated on 22/Aug/21 $$\Omega=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{x}+{tan}\left({sinx}\right)}{\lambda+{cos}\left({x}\right)}{dx} \\…
Question Number 151609 by mathdanisur last updated on 22/Aug/21 Answered by ghimisi last updated on 22/Aug/21 $$\Leftrightarrow{log}_{{xy}} \left(\mathrm{1}+\sqrt{{xy}}\right)^{\mathrm{2}} \geqslant{log}_{\frac{{x}+{y}}{\mathrm{2}}} \left(\frac{{x}+{y}}{\mathrm{2}}+\mathrm{1}\right)\Leftrightarrow \\ $$$$\Leftrightarrow\frac{{ln}\left(\mathrm{1}+\sqrt{{xy}}\right)}{{ln}\sqrt{{xy}}}\geqslant\frac{{ln}\left(\mathrm{1}+\frac{{x}+{y}}{\mathrm{2}}\right)}{{ln}\frac{{x}+{y}}{\mathrm{2}}}\:\:\left(\bullet\right) \\ $$$${f}\left({t}\right)=\frac{{ln}\left(\mathrm{1}+{t}\right)}{{lnt}},{f}:\left(\mathrm{1};\infty\right)\rightarrow{R} \\…
Question Number 151599 by mathdanisur last updated on 22/Aug/21 $$\Omega\:=\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\mathrm{cos}\left(\mathrm{x}^{\boldsymbol{\mathrm{n}}} \right)\:\mathrm{dx}\:=\:? \\ $$ Answered by Lordose last updated on 22/Aug/21 $$ \\ $$$$\Omega\:\overset{\mathrm{x}=\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{n}}}…
Question Number 151596 by mathdanisur last updated on 22/Aug/21 Answered by dumitrel last updated on 22/Aug/21 $$\left(\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}\right)^{\mathrm{2}} \overset{{cbs}} {\leqslant}\mathrm{3}\left({a}+{b}+{c}\right)\Rightarrow\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}\leqslant\sqrt{\mathrm{3}\left({a}+{b}+{c}\right)} \\ $$$$\left(\sqrt{{ab}}+\sqrt{{bc}}+\sqrt{{ac}}\right)^{\mathrm{2}} \overset{{cbs}} {\leqslant}\mathrm{3}\left({ab}+{bc}+{ac}\right)\Rightarrow\sqrt{{ab}}+\sqrt{{bc}}+\sqrt{{ac}}\leqslant\sqrt{\mathrm{3}\left({ab}+{bc}+{ac}\right.} \\ $$$$\Rightarrow\left(\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}\right)\left(\sqrt{{ab}}+\sqrt{{bc}}+\sqrt{{ac}}\leqslant\sqrt{\mathrm{3}\left({a}+{b}+{b}\right)\centerdot\mathrm{3}\left({ab}+{bc}+{ac}\right)}=\right.…
Question Number 86042 by M±th+et£s last updated on 26/Mar/20 $${solve}:\:\:\lfloor\:\sqrt{{x}}\:\rfloor=\lfloor\frac{{x}}{\mathrm{2}}\rfloor \\ $$ Answered by Rio Michael last updated on 26/Mar/20 $$\:\lfloor\sqrt{{x}}\:\rfloor\:=\:\lfloor\frac{{x}}{\mathrm{2}}\rfloor \\ $$$$\:\Rightarrow\:\sqrt{{x}}\:=\:\frac{{x}}{\mathrm{2}} \\ $$$$\:\:\:{x}\:=\:\frac{{x}^{\mathrm{2}}…
Question Number 151573 by amin96 last updated on 22/Aug/21 Answered by Olaf_Thorendsen last updated on 22/Aug/21 $$\mathrm{S}_{{n}} \:=\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{k}+\mathrm{1}\right)\left(\mathrm{3}{k}+\mathrm{2}\right)\left(\mathrm{3}{k}+\mathrm{3}\right)} \\ $$$$\mathrm{S}_{{n}} \:=\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left(\frac{\mathrm{1}/\mathrm{2}}{\mathrm{3}{k}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}{k}+\mathrm{2}}+\frac{\mathrm{1}/\mathrm{2}}{\mathrm{3}{k}+\mathrm{3}}\right)…
Question Number 151560 by peter frank last updated on 21/Aug/21 $$\mathrm{show}\:\mathrm{that}\:\:\mathrm{i}^{\mathrm{i}} \:\:\mathrm{is}\:\mathrm{always}\:\mathrm{real} \\ $$ Answered by puissant last updated on 21/Aug/21 $${i}^{\mathrm{2}} =−\mathrm{1}\:\Rightarrow\:{i}=\sqrt{−\mathrm{1}} \\ $$$${i}^{{i}}…