Question Number 125450 by joki last updated on 11/Dec/20 Answered by Dwaipayan Shikari last updated on 11/Dec/20 $$\int\frac{\mathrm{2}{x}^{\mathrm{3}} \left({x}−\mathrm{1}\right)}{{x}^{\mathrm{2}} −{x}}+\frac{\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{{x}^{\mathrm{2}} −{x}}+\frac{\mathrm{1}}{{x}\left({x}−\mathrm{1}\right)}{dx} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{3}{log}\left({x}\right)+{log}\left(\frac{{x}−\mathrm{1}}{{x}}\right)+{C}…
Question Number 190982 by sciencestudentW last updated on 15/Apr/23 $${p}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{3}} −\mathrm{3}{xy}+\mathrm{10} \\ $$$${a}_{\mathrm{0}} =?\:\:{a}_{\mathrm{1}} =?\:\:\:\:\:{a}_{\mathrm{2}} =?\:\:{a}_{\mathrm{10}} =? \\ $$ Commented by mr W last…
Question Number 125437 by SOMEDAVONG last updated on 11/Dec/20 $$\mathrm{show}\:\mathrm{that}\:\left(\mathrm{E}\right):\mathrm{m}\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2017}} \left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2018}} +\mathrm{2x}+\mathrm{3}=\mathrm{0}\:\mathrm{have}\:\mathrm{root}\:\forall\mathrm{m}\in\mathbb{R}. \\ $$ Commented by Snail last updated on 11/Dec/20 $${I}\:{have}\:{a}\:{doubt}\:{what}\:{type}\:{of}\:{root}\:{it}\:{has}\:{when}\: \\ $$$${m}\in\mathbb{R}….{please}\:{say}\:{the}\:{condition}\:{properly} \\…
Question Number 125426 by joki last updated on 11/Dec/20 Answered by bramlexs22 last updated on 11/Dec/20 $${x}^{\mathrm{2}} \:−\mathrm{1}=\:{u} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{du}}{{u}^{\mathrm{2}} \left({u}+\mathrm{2}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\left[\frac{{A}}{{u}}+\frac{{B}}{{u}^{\mathrm{2}} }+\frac{{C}}{{u}+\mathrm{2}}\right]\:{du} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mid{u}\mid\:−\frac{\mathrm{1}}{\mathrm{2}{u}}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mid{u}+\mathrm{2}\mid\right)+{c} \\…
Question Number 125424 by joki last updated on 11/Dec/20 Answered by bramlexs22 last updated on 11/Dec/20 $$\int\:\frac{{d}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{−\mathrm{2}} {d}\left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{2}\left(\boldsymbol{{x}}+\mathrm{1}\right)}\:+\:\boldsymbol{{c}}\:…
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Question Number 190953 by mustafazaheen last updated on 15/Apr/23 $$\boldsymbol{\mathrm{Hello}}\:\boldsymbol{\mathrm{Tinku}}\:\boldsymbol{\mathrm{Tara}} \\ $$$$\mathrm{When}\:\mathrm{I}\:\mathrm{copy}\:\mathrm{one}\:\mathrm{qustion}\:\mathrm{or}\: \\ $$$$\mathrm{letter}\:\mathrm{from}\:\mathrm{another}\:\mathrm{app}\:\mathrm{I}\:\mathrm{can} \\ $$$$\mathrm{not}\:\mathrm{to}\:\mathrm{paste}\:\mathrm{in}\:\mathrm{this}\:\mathrm{Math}\:\mathrm{Editor} \\ $$$$\mathrm{and}\:\mathrm{copy}\:\mathrm{to}\:\mathrm{Latex}\:\mathrm{do}\:\mathrm{not}\:\mathrm{paste}\: \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{Math}\:\mathrm{Editor} \\ $$ Commented by mustafazaheen…
Question Number 125413 by Hassen_Timol last updated on 11/Dec/20 $$\mathrm{Knowing}\:\mathrm{that}\:: \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}^{\mathrm{2}} \:=\:\begin{pmatrix}{\mathrm{2}{n}}\\{\:\:{n}}\end{pmatrix} \\ $$$$ \\ $$$$\mathrm{We}\:\mathrm{have}\:{p}\:\mathrm{tokens}\:\mathrm{and}\:{n}\:\mathrm{boxes}. \\ $$$$\mathrm{Each}\:\mathrm{box}\:\mathrm{is}\:\mathrm{labeled}\:\mathrm{with}\:\mathrm{a}\:\mathrm{number}\:\mathrm{from}\:\mathrm{1}\:\mathrm{to}\:{n}. \\ $$$$\mathrm{Each}\:\mathrm{box}\:\mathrm{is}\:\mathrm{enough}\:\mathrm{big}\:\mathrm{to}\:\mathrm{receive}\:\mathrm{all}\:{p}\:\mathrm{tokens}. \\ $$$$…
Question Number 59861 by ANTARES VY last updated on 15/May/19 $$\sqrt{\boldsymbol{{a}}+\boldsymbol{{b}}\sqrt{\boldsymbol{{c}}}}=\sqrt{\frac{\boldsymbol{{a}}+\sqrt{\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{c}}}}{\mathrm{2}}}+\sqrt{\frac{\boldsymbol{{a}}−\sqrt{\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{c}}}}{\mathrm{2}}}. \\ $$$$\boldsymbol{{prove}} \\ $$ Commented by Kunal12588 last updated on…
Question Number 59858 by jimful last updated on 15/May/19 $$\mathrm{P}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{{i}}\: \\ $$$$\mathrm{P}_{\mathrm{n}+\mathrm{1}} \mathrm{P}_{\mathrm{n}} =\mathrm{1}−\mathrm{P}_{\mathrm{n}+\mathrm{1}} \\ $$$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}Im}\left(\mathrm{P}_{\mathrm{n}} \right)\:=? \\ $$ Terms of Service Privacy Policy…