Question Number 148157 by aliibrahim1 last updated on 25/Jul/21 Answered by puissant last updated on 26/Jul/21 $$\mathrm{pgcd}\left(\mathrm{a};\mathrm{b}\right)=\mathrm{pgcd}\left(\mathrm{a}−\mathrm{b};\mathrm{b}\right) \\ $$$$\Rightarrow\:\mathrm{pgcd}\left(\mathrm{2}^{\mathrm{a}} −\mathrm{1};\mathrm{2}^{\mathrm{b}} −\mathrm{1}\right)=\mathrm{pgcd}\left(\mathrm{2}^{\mathrm{a}} −\mathrm{2}^{\mathrm{b}} ;\mathrm{2}^{\mathrm{b}} −\mathrm{1}\right) \\…
Question Number 148155 by Khalmohmmad last updated on 25/Jul/21 Answered by Rasheed.Sindhi last updated on 26/Jul/21 $${x}+\frac{\mathrm{1}}{{x}}=−\mathrm{1}\Rightarrow{x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\omega,\omega^{\mathrm{2}} \\ $$$${x}=\omega \\ $$$${x}^{\mathrm{1400}} +{x}^{\mathrm{1363}}…
Question Number 17053 by I’m a gamer last updated on 30/Jun/17 Answered by ajfour last updated on 30/Jun/17 $$\mathrm{L}=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{e}^{\sqrt{\mathrm{A}}} −\mathrm{e}^{\sqrt{\mathrm{B}}} }{\:\sqrt{\mathrm{ln}\:\left(\mathrm{1}+\left(\mathrm{1}−\mathrm{x}\right)\right.}} \\ $$$$\:\:=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{e}^{\sqrt{\mathrm{A}}}…
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Question Number 148105 by abdurehime last updated on 25/Jul/21 Commented by abdurehime last updated on 25/Jul/21 $$\mathrm{help}\:\mathrm{me} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 148082 by tabata last updated on 25/Jul/21 Commented by tabata last updated on 25/Jul/21 $${help}\:{me}\:{msr}\:{olaf} \\ $$ Answered by Olaf_Thorendsen last updated on…
Question Number 148071 by tabata last updated on 25/Jul/21 Commented by tabata last updated on 25/Jul/21 $$??????? \\ $$ Answered by Olaf_Thorendsen last updated on…
Question Number 148068 by tabata last updated on 25/Jul/21 Answered by Olaf_Thorendsen last updated on 25/Jul/21 $$\left(\mathrm{B}\right) \\ $$$$\mathrm{Let}\:{f}\left({z}\right)\:=\:{z}^{\mathrm{7}} −\mathrm{4}{z}^{\mathrm{3}} +{z}−\mathrm{1}\:=\:\mathrm{0},\:\mid{z}\mid\:=\:\mathrm{1} \\ $$$$\mathrm{and}\:{g}\left({z}\right)\:=\:−\mathrm{4}{z}^{\mathrm{3}} \\ $$$$…
Question Number 148064 by tabata last updated on 25/Jul/21 Commented by tabata last updated on 25/Jul/21 $${help}\:{me}\:{sir}\:{in}\:{complex}\:{number} \\ $$ Answered by Olaf_Thorendsen last updated on…
Question Number 148070 by tabata last updated on 25/Jul/21 Commented by tabata last updated on 25/Jul/21 $$????? \\ $$ Terms of Service Privacy Policy Contact:…