a)
\(\dfrac{2a^2-3a-2}{a^2-4}=2\left(a\ne\pm2\right)\\ \Leftrightarrow2a^2-3a-2=2\left(a^2-4\right)\\ \Leftrightarrow2a^2-3a-2=2a^2-8\\ \Leftrightarrow-3a-2=-8\\ \Leftrightarrow-3a=-8+2=-6\\ \Leftrightarrow a=\dfrac{-6}{-3}=2\left(ktm\right)\)
=> không có a thỏa mãn
b)
\(\dfrac{3a-1}{3a+1}+\dfrac{a-3}{a+3}=2\left(a\ne-\dfrac{1}{3};a\ne-3\right)\\ \Leftrightarrow\dfrac{\left(3a-1\right)\left(a+3\right)+\left(a-3\right)\left(3a+1\right)}{\left(3a+1\right)\left(a+3\right)}=2\\ \Leftrightarrow3a^2+9a-a-3+3a^2-9a+a-3=2\left(3a+1\right)\left(a+3\right)\\ \Leftrightarrow6a^2-6=2\left(3a^2+10a+3\right)\\ \Leftrightarrow6a^2-6=6a^2+20a+6\\ \Leftrightarrow20a+6=-6\\ \Leftrightarrow20a=-12\\ \Leftrightarrow a=-\dfrac{12}{20}=-\dfrac{3}{5}\left(tm\right)\)
