If frac{1}{(x-1)(x-2)(x-3)}=frac{A}{x-1}+frac | vedclass

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(A) Given $\frac{1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$.Multiplying by $(x-1)(x-2)(x-3)$, we get $1=A(x-2)(x-3)+B(x-1)(x-3)+C(

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$P+Q+R$

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(A) Given $\frac{1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$.Multiplying by $(x-1)(x-2)(x-3)$, we get $1=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)$.For $x=1$, $1=A(1-2)(1-3) \Rightarrow A=\frac{1}{2}$.For $x=2$, $1=B(2-1)(2-3) \Rightarrow B=-1$.For $x=3$, $1=C(3-1)(3-2) \Rightarrow C=\frac{1}{2}$.Thus, $A+2B+3C = \frac{1}{2} + 2(-1) + 3(\frac{1}{2}) = \frac{1}{2} - 2 + \frac{3}{2} = 0$.Now, for $\frac{x}{(x-1)(x-2)(x-3)}=\frac{P}{x-1}+\frac{Q}{x-2}+\frac{R}{x-3}$, we have $x=P(x-2)(x-3)+Q(x-1)(x-3)+R(x-1)(x-2)$.For $x=1$, $1=P(1-2)(1-3) \Rightarrow P=\frac{1}{2}$.For $x=2$, $2=Q(2-1)(2-3) \Rightarrow Q=-2$.For $x=3$, $3=R(3-1)(3-2) \Rightarrow R=\frac{3}{2}$.Calculating $P+Q+R = \frac{1}{2} - 2 + \frac{3}{2} = 0$.Since both expressions equal $0$, $A+2B+3C = P+Q+R$.

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 $A+2B+3C=$

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