If $\frac{x^2-2}{(x^2+1)(x^2+3)} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+3}$, then $D=$

If we resolve the rational fraction $\frac{1}{(1-2x)^2(1-3x)}$ into partial fractions of the form $\frac{A}{1-3x} + \frac{B}{1-2x} + \frac{C}{(1-2x)^2}$,then what is $\min \{A, B, C\} = $

$\begin{aligned} & \text{If } \frac{4x^2+5x^4+7}{(x^2+1)(x^4+x^2+1)} = \frac{Ax+B}{x^2+1} \\ & + \frac{Cx^3+Dx^2+Ex+F}{x^4+x^2+1}, \text{ then } \\ & B+2(D+F+E)-C \cdot A = \end{aligned}$

If $\frac{8}{(x+3)^2(x-2)}=\frac{Ax+B}{(x+3)^2}+\frac{C}{x-2}$,then $25(B+8C-A)=$

If $\frac{3x^2+x+1}{(x-1)^4} = \frac{a}{(x-1)} + \frac{b}{(x-1)^2} + \frac{c}{(x-1)^3} + \frac{d}{(x-1)^4}$,then $\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ is equal to

For $|x| < 1$,the coefficient of $x^2$ in the power series expansion of $\frac{x^4}{(x+1)(x-2)}$ is