यदि $\frac{ax+5}{(x^2+b)(x+3)}=\frac{x+21}{12(x^2+b)}+\frac{c}{12(x+3)}$ है,तो $b^2=$

यदि $\frac{1}{(3x+1)(x-2)}=\frac{A}{3x+1}+\frac{B}{x-2}$ और $\frac{x+1}{(3x+1)(x-2)}=\frac{C}{3x+1}+\frac{D}{x-2}$ है,तो

यदि $\frac{A}{x-a}+\frac{B x+C}{x^2+b^2}=\frac{1}{(x-a)(x^2+b^2)}$ है,तो $C=$

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

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

यदि $\frac{1}{x(x + 1)(x + 2)...(x + n)} = \frac{A_0}{x} + \frac{A_1}{x + 1} + \frac{A_2}{x + 2} + .... + \frac{A_n}{x + n}$ है,तो $A_r = $