If $\frac{(x - a)(x - b)}{(x - c)(x - d)} = \frac{A}{x - c} - \frac{B}{x - d} + C$,then $C = $

The coefficient of $x^n$ in the expression $\frac{5x + 6}{(2 + x)(1 - x)}$ when expanded in ascending order of $x$ is:

For any quadratic polynomial $f(x)$, it is true that $f(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2!}(x-a)^2$ where $a$ is any real number. If $\frac{3 x^2+4 x+7}{(x-2)^3}=\frac{A}{(x-2)^3}+\frac{B}{(x-2)^2}+\frac{C}{(x-2)}$ and $g(x)=3 x^2+4 x+7$, then $A+B+C=$

If $\frac{3 x^4+5 x^2+2}{\left(x^2+1\right)^2\left(x^2+2\right)}=\frac{A x+B}{x^2+2}+\frac{C x+D}{x^2+1}+\frac{E x+F}{\left(x^2+1\right)^2}$,then $A+2 B+D+4 E=$

Let $a, b$, and $c$ be such that $\frac{1}{(1-x)(1-2x)(1-3x)} = \frac{a}{1-x} + \frac{b}{1-2x} + \frac{c}{1-3x}$. Then $\frac{a}{1} + \frac{b}{3} + \frac{c}{5}$ is equal to:

$\begin{aligned} & \text{If } \frac{x^4}{(x-a)(x-b)(x-c)}=P(x)+\frac{A}{x-a}+\frac{B}{x-b} \\ & +\frac{C}{x-c} \text{, then } P(0)+A(a-b)(a-c)= \end{aligned}$