यदि $\frac{(x+1)}{(2 x-1)(3 x+1)}=\frac{A}{(2 x-1)}+\frac{B}{(3 x+1)}$ है,तो $16 A+9 B$ का मान ज्ञात कीजिए।

  • A
    $4$
  • B
    $5$
  • C
    $6$
  • D
    $8$

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यदि $\frac{2 x+7}{\left(x^2+4\right)\left(x^2+9\right)\left(x^2+16\right)}=\frac{A x+1}{x^2+4}+\frac{B x+m}{x^2+9}+\frac{C x+n}{x^2+16}$ है,तो $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=$

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$\frac{x^2-1}{(x^2+1)(x^2+2)}$ के घात श्रेणी विस्तार में $x^4$ का गुणांक है

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