Question Number 226721 by Spillover last updated on 11/Dec/25 Answered by breniam last updated on 12/Dec/25 $${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xyy}'=\mathrm{0} \\ $$$${t}=\mathrm{ln}\mid{x}\mid \\ $$$${x}={e}^{{t}} \\ $$$${y}\left({x}\right)={z}\left({t}\right)…
Question Number 226697 by Linton last updated on 10/Dec/25 Commented by Linton last updated on 10/Dec/25 $${The}\:{letters}\:{in}\:{TWO}\:{UV}\:{PAIRS}\: \\ $$$${have}\:{the}\:{values}\:\mathrm{0}\:\mathrm{1}\:\mathrm{2}…\:\mathrm{9}\:{in}\:{some} \\ $$$${order}\:{with}\:{each}\:{letter}\:{represent} \\ $$$${ing}\:{a}\:{different}\:{digit}. \\ $$…
Question Number 226612 by Spillover last updated on 07/Dec/25 Commented by SonGoku last updated on 07/Dec/25 $$\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{system}}\:\begin{cases}{\boldsymbol{\mathrm{m}}\:+\:\boldsymbol{\mathrm{m}}\:=\:\mathrm{20}}\\{\boldsymbol{\mathrm{c}}\:−\:\boldsymbol{\mathrm{l}}\:=\:\mathrm{3}}\\{\boldsymbol{\mathrm{m}}\:−\:\boldsymbol{\mathrm{l}}×\boldsymbol{\mathrm{c}}\:=\:?}\end{cases}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 226609 by Spillover last updated on 07/Dec/25 Answered by Ghisom_ last updated on 08/Dec/25 $$\underset{−\mathrm{2}} {\overset{−\mathrm{1}} {\int}}\frac{\sqrt{\mathrm{2}+{x}}}{{x}\sqrt{\mathrm{2}−{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\sqrt{\mathrm{2}+{x}}}{\:\sqrt{\mathrm{2}−{x}}}\:\rightarrow\:{dx}=\frac{\sqrt{\left(\mathrm{2}−{x}\right)^{\mathrm{3}} \left(\mathrm{2}+{x}\right)}}{\mathrm{2}}{dt}\right] \\ $$$$=\mathrm{4}\underset{\mathrm{0}} {\overset{\mathrm{1}/\sqrt{\mathrm{3}}}…
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Question Number 226513 by aba_math last updated on 01/Dec/25 $${Find}\:{gcd}\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} ,{ab}\right)\:{if}\:{gcd}\left({a},{b}\right)=\mathrm{1} \\ $$ Answered by peace2 last updated on 02/Dec/25 $${a}={dx};{b}={dy};{gcd}\left({x},{y}\right)=\mathrm{1} \\ $$$${gcd}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}}…
Question Number 226455 by Spillover last updated on 29/Nov/25 Answered by Ghisom_ last updated on 29/Nov/25 $$\frac{\mathrm{2}−{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}=\frac{\mathrm{1}+\mathrm{1}−{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)\sqrt{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}}= \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}/\mathrm{2}} \left(\mathrm{1}+{x}\right)^{\mathrm{1}/\mathrm{2}} }+\frac{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}/\mathrm{2}} \left(\mathrm{1}+{x}\right)^{\mathrm{1}/\mathrm{2}}…
Question Number 226464 by Spillover last updated on 29/Nov/25 Commented by Frix last updated on 29/Nov/25 $$\mathrm{By}\:\mathrm{parts} \\ $$$${u}'=\mathrm{1}\:\rightarrow\:{u}={x} \\ $$$${v}=\mathrm{cot}^{−\mathrm{1}} \:\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:\rightarrow\:{v}'=−\frac{\mathrm{2}{x}−\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)}…