Question Number 211986 by Spillover last updated on 26/Sep/24 Answered by Spillover last updated on 28/Sep/24 Answered by Spillover last updated on 28/Sep/24 Commented by…
Question Number 212006 by RojaTaniya last updated on 26/Sep/24 Answered by a.lgnaoui last updated on 26/Sep/24 $$\:\:\mathrm{posons}\:\:\boldsymbol{\mathrm{z}}=\boldsymbol{\mathrm{x}}+\mathrm{100} \\ $$$$\:\:\frac{\left(\boldsymbol{\mathrm{z}}−\mathrm{2}\right)^{\mathrm{5}} +\left(\boldsymbol{\mathrm{z}}+\mathrm{2}\right)^{\mathrm{5}} }{\left(\boldsymbol{\mathrm{z}}−\mathrm{1}\right)^{\mathrm{5}} +\left(\boldsymbol{\mathrm{z}}+\mathrm{1}\right)^{\mathrm{5}} }=\frac{\mathrm{2}\boldsymbol{\mathrm{z}}^{\mathrm{5}} +\mathrm{80}\boldsymbol{\mathrm{z}}^{\mathrm{3}} +\mathrm{160}\boldsymbol{\mathrm{z}}}{\mathrm{2}\boldsymbol{\mathrm{z}}^{\mathrm{5}}…
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Question Number 211949 by Spillover last updated on 25/Sep/24 Answered by MathematicalUser2357 last updated on 26/Sep/24 $$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{the}\:\mathrm{integration}\:\mathrm{idea}! \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{\mathrm{1}}{{x}}\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\mathrm{ln}\left(\frac{{ax}^{\mathrm{2}} +\sqrt{\mathrm{2}{a}}{x}+\mathrm{1}}{{ax}^{\mathrm{2}} −\sqrt{\mathrm{2}{a}}{x}+\mathrm{1}}\right){dx}=\mathrm{4}\pi\:\mathrm{cot}^{−\mathrm{1}} \sqrt{\frac{\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}+\mathrm{1}}{{a}}}…
Question Number 211946 by BaliramKumar last updated on 25/Sep/24 Answered by BHOOPENDRA last updated on 25/Sep/24 $$\left(\mathrm{2}+\mathrm{3}+\mathrm{4}\right)^{\mathrm{2}} =\mathrm{9}^{\mathrm{2}} =\mathrm{81} \\ $$ Commented by BHOOPENDRA last…
Question Number 211954 by RojaTaniya last updated on 25/Sep/24 $$\:{x},{y}\:{are}\:{rational}\:{numbers}\:{where} \\ $$$$\:{x}\neq\mathrm{0},\:{y}\neq\mathrm{0},\:{x}\neq{y},\:{then}\:{is}\:{it}\: \\ $$$$\:{possible}:\:\:{x}^{\mathrm{5}} +{y}^{\mathrm{5}} =\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \:? \\ $$ Answered by Frix last updated…
Question Number 211932 by a.lgnaoui last updated on 25/Sep/24 $$\mathrm{determiner}\:\:\:\:\boldsymbol{\mathrm{R}}\mathrm{1}\:\:\:\:\boldsymbol{\mathrm{R}}\mathrm{2}\:\mathrm{et}\:\boldsymbol{\mathrm{R}}\mathrm{3} \\ $$$$\mathrm{segment}\:\mathrm{de}\:\mathrm{longueur}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{est}}\:\boldsymbol{\mathrm{tangent}}\:\boldsymbol{\mathrm{aux}} \\ $$$$\boldsymbol{\mathrm{cercles}}\:\mathrm{1}\boldsymbol{\mathrm{et}}\:\mathrm{2}. \\ $$$$\:\boldsymbol{\mathrm{MN}}//\boldsymbol{\mathrm{EF}};\:\:\boldsymbol{\mathrm{EF}}=\boldsymbol{\mathrm{a}};\:\:\mathrm{OM}=\mathrm{ON}=\frac{\mathrm{3}\boldsymbol{\mathrm{a}}}{\mathrm{2}}. \\ $$$$\left(\mathrm{length}\:\boldsymbol{\mathrm{a}}\:\mathrm{is}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{cirles}\:\mathrm{C1}\:\left(\mathrm{radius}\:\mathrm{R1}\right)\mathrm{and}\:\mathrm{circldC2}\left(\mathrm{radius}\:\mathrm{R2}\right)\right). \\ $$ Commented by a.lgnaoui last updated…
Question Number 211918 by Spillover last updated on 24/Sep/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 211917 by Spillover last updated on 24/Sep/24 Answered by BHOOPENDRA last updated on 24/Sep/24 $${Use}\:{L}'{Hopital}'{s}\:{Rule}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{mn}\left(\mathrm{1}+{nx}\right)^{{m}−\mathrm{1}} −{mn}\left(\mathrm{1}+{mx}\right)^{{n}−\mathrm{1}} }{\left(\mathrm{1}+{mx}\right)^{\frac{\mathrm{1}}{{m}}−\mathrm{1}} −\left(\mathrm{1}+{nx}\right)^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} } \\…
Question Number 211919 by Spillover last updated on 24/Sep/24 Answered by Frix last updated on 24/Sep/24 $${x}^{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{4}}{\mathrm{4}}+\frac{\mathrm{5}}{\mathrm{8}}+\frac{\mathrm{6}}{\mathrm{16}}…} ={x}^{\mathrm{5}} \\ $$$$\mathrm{1}+\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{k}+\mathrm{2}}{\mathrm{2}^{{k}} }\:=\mathrm{1}+\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{2}^{{k}}…