The partial fraction decomposition of $\frac{9x-7}{(x+3)(x^2+1)}$ is

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:

Let $H(x)=3x^4+6x^3-2x^2+1$ and $g(x)$ be a polynomial of degree one. If $\frac{H(x)}{(x-1)(x+1)(x-2)}=f(x)+\frac{g(x)}{(x-1)(x+1)(x-2)}$, then $H(-1)+2H(2)-3H(1)=$

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

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 $\frac{3x + 4}{x^2 - 3x + 2} = \frac{A}{x - 2} - \frac{B}{x - 1}$,then $(A, B) = $