6. For any $x \in\left(-\frac{1}{2}, 1\right)$, we have
$\frac{x}{1+x-2 x^{2}}=a_{0}+a_{1} x+a_{2} x^{2}+\square+a_{n} x^{n}+\square$, then the value of $a_{3}+a_{4}$ is $\qquad$
6. -2

Analysis: Given $a_{0}=0$, hence $\frac{1}{1+x-2 x^{2}}=a_{1}+a_{2} x+\square+a_{n} x^{n-1}+\square$, $1=\left(1+x-2 x^{2}\right)\left(a_{1}+a_{2} x+\mathrm{L}+a_{n} x^{n-1}+\mathrm{L}\right)$,

thus $a_{1}=1$, then $1=\left(1+x-2 x^{2}\right)\left(1+a_{2} x+\mathrm{L}+a_{n} x^{n-1}+\mathrm{L}\right)$.

Expanding and comparing the coefficients of the first and third terms, we get: $a_{2}+1=0, a_{4}+a_{3}-2 a_{2}=0$, hence $a_{4}+a_{3}=-2$.