If frac{3x^2+x+1}{(x-1)^4} = frac{a}{(x-1)} + | vedclass

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(B) Let $x-1 = t$,so $x = t+1$. Substituting this into the numerator: $3(t+1)^2 + (t+1) + 1 = 3(t^2+2t+1) + t + 2 = 3t^2 + 6t + 3 + t + 2 = 3t^2 + 7t

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$\left[\begin{array}{ll}0 & 3 \ 7 & 5\end{array}ight]$

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(B) Let $x-1 = t$,so $x = t+1$. Substituting this into the numerator: $3(t+1)^2 + (t+1) + 1 = 3(t^2+2t+1) + t + 2 = 3t^2 + 6t + 3 + t + 2 = 3t^2 + 7t + 5$. Now,the expression becomes: $\frac{3t^2+7t+5}{t^4} = \frac{3}{t^2} + \frac{7}{t^3} + \frac{5}{t^4}$. Comparing this with $\frac{a}{t} + \frac{b}{t^2} + \frac{c}{t^3} + \frac{d}{t^4}$,we get: $a = 0, b = 3, c = 7, d = 5$. Thus,$\left[\begin{array}{ll}a & b \ c & d\end{array}ight] = \left[\begin{array}{ll}0 & 3 \ 7 & 5\end{array}ight]$.

If $\frac{3x^2+x+1}{(x-1)^4} = \frac{a}{(x-1)} + \frac{b}{(x-1)^2} + \frac{c}{(x-1)^3} + \frac{d}{(x-1)^4}$,then $\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ is equal to

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