Category: Arithmetic

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}}}…

Question Number 226585 by Spillover last updated on 06/Dec/25 Terms of Service Privacy Policy Contact: info@tinkutara.com

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 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)}…

Question Number 226006 by Linton last updated on 18/Nov/25 $$\left(\mathrm{3}/\mathrm{7}\right)^{\mathrm{0}} \:\:\:{prove}\:{and}\:{evalute}\:{show}\:{all} \\ $$$${working} \\ $$ Commented by mr W last updated on 18/Nov/25 $${for}\:{any}\:{a}\neq\mathrm{0}\:{we}\:{have} \\…