{2x+a,x⩽0ln(1+2x),x>0\begin{cases}2x + a, & x\leqslant0\\\ln(1 + 2x), & x>0\end{cases}{2x+a,ln(1+2x),x⩽0x>0在点x=0x = 0x=0处连续,求aaa的值.
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