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If $\frac{3 \mathrm{x}+2}{(\mathrm{x}+1)\left(2 \mathrm{x}^2+3\right)}=\frac{\mathrm{A}}{\mathrm{x}+1}+\frac{\mathrm{Bx}+\mathrm{C}}{2 \mathrm{x}^2+3}$, then $\mathrm{A}-\mathrm{B}+\mathrm{C}=$
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The correct answer is:
$2$
$\frac{3 x+2}{(x+1)\left(2 x^2+3\right)}=\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B} x+\mathrm{C}}{2 x^2+3}$
$\begin{aligned} & (3 x+2)=\mathrm{A}\left(2 x^2+3\right)+(x+1)(\mathrm{B} x+\mathrm{C}) \\ & (3 x+2)=\mathrm{A} \cdot 2 x^2+3 \mathrm{~A}+\mathrm{B} x^2+\mathrm{C} x+\mathrm{B} x+\mathrm{C} \\ & =(2 \mathrm{~A}+\mathrm{B}) x^2+(\mathrm{B}+\mathrm{C}) x+(3 \mathrm{~A}+\mathrm{C})\end{aligned}$
$2 A+B=0$...(i)
$B+C=3$...(ii)
$3 A+C=2$...(iii)
From (i), (ii) and (iii)
$$
\begin{aligned}
& A=\frac{-1}{5}, B=\frac{2}{5}, C=\frac{13}{5} \\
& A-B+C=\frac{-1}{5}-\frac{2}{5}+\frac{13}{5}=2
\end{aligned}
$$
$\begin{aligned} & (3 x+2)=\mathrm{A}\left(2 x^2+3\right)+(x+1)(\mathrm{B} x+\mathrm{C}) \\ & (3 x+2)=\mathrm{A} \cdot 2 x^2+3 \mathrm{~A}+\mathrm{B} x^2+\mathrm{C} x+\mathrm{B} x+\mathrm{C} \\ & =(2 \mathrm{~A}+\mathrm{B}) x^2+(\mathrm{B}+\mathrm{C}) x+(3 \mathrm{~A}+\mathrm{C})\end{aligned}$
$2 A+B=0$...(i)
$B+C=3$...(ii)
$3 A+C=2$...(iii)
From (i), (ii) and (iii)
$$
\begin{aligned}
& A=\frac{-1}{5}, B=\frac{2}{5}, C=\frac{13}{5} \\
& A-B+C=\frac{-1}{5}-\frac{2}{5}+\frac{13}{5}=2
\end{aligned}
$$
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