Question Number 195446 by sonukgindia last updated on 02/Aug/23 Answered by Frix last updated on 02/Aug/23 $$\sqrt{{a}+{b}\mathrm{i}}=\frac{\sqrt{{a}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}}{\:\sqrt{\mathrm{2}}}+\mathrm{sign}\:{b}\:\frac{\sqrt{−{a}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}}{\:\sqrt{\mathrm{2}}}\mathrm{i} \\ $$$$\sqrt{−\mathrm{110}+\mathrm{66i}}=\sqrt{\mathrm{11}\left(−\mathrm{5}+\sqrt{\mathrm{34}}\right)}+\mathrm{i}\sqrt{\mathrm{11}\left(\mathrm{5}+\sqrt{\mathrm{34}}\right)} \\ $$…
Question Number 195414 by tri26112004 last updated on 02/Aug/23 $${x}^{{n}} +\frac{\mathrm{1}}{{x}^{{n}} }=¿ \\ $$ Answered by Frix last updated on 02/Aug/23 $$=\mathrm{2cosh}\:\left({n}\mathrm{ln}\:{x}\right) \\ $$ Commented…
Question Number 195436 by MrGHK last updated on 02/Aug/23 $$\underset{\boldsymbol{\mathrm{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\boldsymbol{\mathrm{a}}\right)^{−\boldsymbol{\mathrm{n}}} }{\left(\boldsymbol{\mathrm{n}}+\mathrm{2}\right)^{\mathrm{2}} }\left(\boldsymbol{\psi}\left(\frac{\boldsymbol{\mathrm{n}}+\mathrm{4}}{\mathrm{2}}\right)−\boldsymbol{\psi}\left(\frac{{n}+\mathrm{3}}{\mathrm{2}}\right)\right)=??? \\ $$ Answered by witcher3 last updated on 02/Aug/23 $$\Psi\left(\mathrm{1}+\mathrm{x}\right)=−\gamma+\int_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 195411 by sonukgindia last updated on 02/Aug/23 Answered by MM42 last updated on 02/Aug/23 $${lim}_{{x}\rightarrow\infty} \:\frac{\mathrm{2}×\mathrm{2}^{{x}} −\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{9}^{{x}} +\mathrm{4}^{{x}} }{\mathrm{8}×\mathrm{4}^{{x}} +\mathrm{3}^{{x}} −\frac{\mathrm{1}}{\mathrm{4}}×\mathrm{4}^{{x}} } \\…
Question Number 195433 by sonukgindia last updated on 02/Aug/23 Answered by gatocomcirrose last updated on 02/Aug/23 $$\mathrm{2}\begin{vmatrix}{\mathrm{2}}&{−\mathrm{2}}&{\mathrm{1}}\\{\mathrm{x}}&{\mathrm{2}}&{−\mathrm{1}}\\{\mathrm{1}}&{−\mathrm{1}}&{−\mathrm{2}}\end{vmatrix}−\mathrm{3}\begin{vmatrix}{\mathrm{x}}&{−\mathrm{2}}&{\mathrm{1}}\\{\mathrm{2}}&{\mathrm{2}}&{−\mathrm{1}}\\{\mathrm{x}}&{−\mathrm{1}}&{−\mathrm{2}}\end{vmatrix}+ \\ $$$$+\mathrm{i}\begin{vmatrix}{\mathrm{x}}&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{2}}&{\mathrm{x}}&{−\mathrm{1}}\\{\mathrm{x}}&{\mathrm{1}}&{−\mathrm{2}}\end{vmatrix}+\begin{vmatrix}{\mathrm{x}}&{\mathrm{2}}&{−\mathrm{2}}\\{\mathrm{2}}&{\mathrm{x}}&{\mathrm{2}}\\{\mathrm{x}}&{\mathrm{1}}&{−\mathrm{1}}\end{vmatrix}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}\left[−\mathrm{10}−\mathrm{5x}\right]−\mathrm{3}\left[−\mathrm{5x}−\mathrm{10}\right]+\mathrm{i}\left[−\mathrm{3x}^{\mathrm{2}} −\mathrm{x}+\mathrm{10}\right]+\left[\mathrm{x}^{\mathrm{2}} +\mathrm{2x}\right]=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}}…
Question Number 195385 by Rodier97 last updated on 01/Aug/23 $$ \\ $$$$ \\ $$$$\mathrm{lim}_{{x}\Rightarrow\mathrm{a}^{+} } \:\:\:\:\frac{\sqrt{{x}}\:−\sqrt{\mathrm{a}}\:−\sqrt{{x}−\mathrm{a}}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}\:\:;\:\:\mathrm{a}\:>\:\mathrm{0} \\ $$ Answered by cortano12 last updated…
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Question Number 195342 by Matica last updated on 31/Jul/23 $$\:\:\mathrm{1}.\:\mathrm{Prove}\:\mathrm{that}\:\:\forall{n}\:\in\:\mathbb{N}^{\ast} \:,\:\mathrm{4}^{{n}} \left({n}!\right)^{\mathrm{3}} \:<\:\left({n}+\mathrm{1}\right)^{\mathrm{3}{n}} \:. \\ $$$$\mathrm{2}.\:\mathrm{Solve}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{in}\:\mathbb{Z}^{\mathrm{2}} \:: \\ $$$$\:\:\:\:\:{a}./\:\:\mathrm{2}{x}^{\mathrm{3}} +{xy}−\mathrm{7}=\mathrm{0}\:, \\ $$$$\:\:\:\:\:{b}./\:\:{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{7}\right)\left({x}+\mathrm{8}\right)={y}^{\mathrm{2}} . \\ $$…
Question Number 195343 by Matica last updated on 31/Jul/23 $$\:{hello}\:{everyone},\:{I}'{m}\:{a}\:{new}\:{student}. \\ $$$$\:{I}\:{beg}\:{your}\:{pardon}. \\ $$$$\:{Please}\:{tell}\:{me}\:{any}\:{good}\:{math}\:{book} \\ $$$$\:{you}\:{know}.\:{Thanks}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com