Question

Exercițiul 2 din imagine!​

Answer 1

 

$\displaystyle\bf\\\frac{2a+i}{1-i}~~ar~fi~numar~real~daca:\\\\\textbf{Raportul dintre partea reala de la numarator si partea reala}\\\textbf{de la numitor ar fi egal cu raportul partilor imaginare, }\\\textbf{astfel incat sa putem da un factor comun la numarator }\\\textbf{si sa se simplifice } (1-i)~cu~(1-i).$

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$\displaystyle\bf\\Rezolvare:\\\\\frac{+i}{-i}=-1\\\\\implies~\frac{2a}{1}=-1\\\\2a=-1\\\\\boxed{\bf a=-\frac{1}{2}}\\\\Verificare:\\\\ \frac{2a+i}{1-i}=\frac{\Big(2\times\Big(-\dfrac{1}{2}\Big)\Big)+i}{1-i}=\frac{\Big(-\dfrac{2}{2}\Big)+i}{1-i}=\frac{-1+i}{1-i}=\\\\(\textbf{La numarator dam factor comun pe }~-1)\\\\=\frac{-1(1-i)}{(1-i)}=\boxed{\bf-1\in R}\\\\\textbf{Am simplificat fractia cu }~(1-i)$