Embibe Experts Solutions for Chapter: Partial Fractions, Exercise 1: TS EAMCET 2018 (04-May Shift-1)

If $\frac{2x+7}{\left(x^2+4\right)\left(x^2+9\right)\left(x^2+16\right)}=\frac{Ax+1}{x^2+4}+\frac{Bx+m}{x^2+9}+\frac{Cx+n}{x^2+16},$ then $\frac{1}{ A}+\frac{1}{ B}+\frac{1}{C}=$

If $\frac{1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3},$ and $\frac{x}{(x-1)(x-2)(x-3)}=\frac{P}{x-1}+\frac{Q}{x-2}+\frac{R}{x-3}$ then $A+2B+3C=$

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

If the partial fraction decomposition of $\frac{x^2+1}{x^3+3x^2+3x+2}$ is $\frac{A}{x+2}+\frac{B}{x^2+x+1}+\frac{C}{\left(x+2\right)\left(x^2+x+1\right)}$, then $A-B+C=$

If $\frac{9x-7}{\left(x+3\right)\left(x^2+1\right)}=\frac{A}{x+3}+\frac{Bx+C}{x^2+1}$, where $A, B, C\in ℝ,$ $A+B+C=$

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