[tex]-|a|-|-a|+2|a+1|-|3a|-2 = -3a \\ \\ -|a|-|a|+2|a+1|-|3|\cdot |a|-2 = -3a \\ \\ -|a|\cdot (1+1) +2|a+1|-3|a|-2 = -3a \\ \\ -2|a|-3|a|+2|a+1|-2 = -3a \\\\ $ $ a\ \textgreater \ 4 \Rightarrow a\rightarrow pozitiv.\\ a\ \textgreater \ 4\Big|+1 \Rightarrow a+1\ \textgreater \ 5 \Rightarrow a+1\rightarrow pozitiv. \\ $(putem elimina modulul)\\ $ \\ -2a-3a+2(a+1)-2 = -3a \\ \\ -5a+2a+2-2 = -3a \\ \\ -3a = -3a \quad $(A)$[/tex]
Acolo era -2 ci nu -8.
