Question Number 211979 by Spillover last updated on 25/Sep/24 Answered by BHOOPENDRA last updated on 25/Sep/24 $$\int\frac{{dx}}{\left({x}^{\mathrm{2}} \mathrm{tan}^{−\mathrm{1}} {x}+\mathrm{tan}^{−\mathrm{1}} {x}\:+{x}^{\mathrm{2}} \pi+\pi\right)} \\ $$$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{tan}^{−\mathrm{1}} {x}+\pi\right)}…
Question Number 211956 by Durganand last updated on 25/Sep/24 Answered by Frix last updated on 25/Sep/24 $$\mathrm{tan}\:\alpha\:={t}\:\:\:\:\:\mathrm{tan}\:\mathrm{2}\alpha\:=\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }\:\:\:\:\:\mathrm{sin}\:\mathrm{2}\alpha\:=\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{{t}}−\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{2}{t}}=\frac{\mathrm{2}−\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\mathrm{2}{t}}=\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{2}{t}}=\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{2}\alpha} \\…
Question Number 211943 by Frix last updated on 25/Sep/24 $$\mathrm{2}^{{m}−\mathrm{1}} =\mathrm{1}+{mn} \\ $$$${m},\:{n}\:\in\mathbb{Z} \\ $$ Commented by BHOOPENDRA last updated on 25/Sep/24 $$\left\{\left({m},{n}\right)\in\mathbb{Z}×\mathbb{Z}\:\mid\:{m}\:{is}\:{odd}\:\&{n}=\frac{\mathrm{2}^{{m}−\mathrm{1}} −\mathrm{1}}{{m}}\right\} \\…
Question Number 211953 by efronzo1 last updated on 25/Sep/24 Commented by BHOOPENDRA last updated on 25/Sep/24 $${h}=\frac{\sqrt{\mathrm{2}}\:\left(\sqrt{{ab}}\:−{a}\:+{b}−{c}\right)}{\:\sqrt{\left(\sqrt{{ab}}\right)+{b}}} \\ $$ Commented by BHOOPENDRA last updated on…
Question Number 211954 by RojaTaniya last updated on 25/Sep/24 $$\:{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 211917 by Spillover last updated on 24/Sep/24 Answered by BHOOPENDRA last updated on 24/Sep/24 $${Use}\:{L}'{Hopital}'{s}\:{Rule}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{mn}\left(\mathrm{1}+{nx}\right)^{{m}−\mathrm{1}} −{mn}\left(\mathrm{1}+{mx}\right)^{{n}−\mathrm{1}} }{\left(\mathrm{1}+{mx}\right)^{\frac{\mathrm{1}}{{m}}−\mathrm{1}} −\left(\mathrm{1}+{nx}\right)^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} } \\…
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Question Number 211919 by Spillover last updated on 24/Sep/24 Answered by Frix last updated on 24/Sep/24 $${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}} \\ $$$$\mathrm{1}+\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{k}+\mathrm{2}}{\mathrm{2}^{{k}} }\:=\mathrm{1}+\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{2}^{{k}}…
Question Number 211920 by Spillover last updated on 24/Sep/24 Answered by aleks041103 last updated on 24/Sep/24 $$\sqrt{{x}\sqrt{{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{3}} \sqrt{…}}}\:}=\:{x}^{\mathrm{1}/\mathrm{2}} {x}^{\mathrm{2}/\mathrm{4}} {x}^{\mathrm{3}/\mathrm{8}} …{x}^{{n}/\mathrm{2}^{{n}} } …\:= \\…