If 3x2+x+1(x−1)4=Ax−1+B(x−1)2+C(x−1)3+D(x−1)4, then (C)A

Put x−1=t⇒x=t+1 3(t+1)2+(t+1)+1t4=3t2+7t+5t4=3t2+7t3+5t4A=0,B=3,C=7,D=5

If 3x2+x+1(x−1)4=Ax−1+B(x−1)2+C(x−1)3+D(x−1)4, then (C)A

Put x−1=t⇒x=t+1 3(t+1)2+(t+1)+1t4=3t2+7t+5t4=3t2+7t3+5t4A=0,B=3,C=7,D=5

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If $\frac{3 x^{2} + x + 1}{{(x - 1)}^{4}} = \frac{A}{x - 1} + \frac{B}{{(x - 1)}^{2}} + \frac{C}{{(x - 1)}^{3}} + \frac{D}{{(x - 1)}^{4}},$ then ${(C)}^{A}$

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Put $x - 1 = t \Rightarrow x = t + 1$
$\begin{array}{l} \frac{3 {(t + 1)}^{2} + (t + 1) + 1}{t^{4}} = \frac{3 t^{2} + 7 t + 5}{t^{4}} = \frac{3}{t^{2}} + \frac{7}{t^{3}} + \frac{5}{t^{4}} \\ A = 0, B = 3, C = 7, D = 5 \end{array}$

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If 3x2+x+1(x−1)4=Ax−1+B(x−1)2+C(x−1)3+D(x−1)4, then (C)A

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If $\frac{3 x^{2} + x + 1}{{(x - 1)}^{4}} = \frac{A}{x - 1} + \frac{B}{{(x - 1)}^{2}} + \frac{C}{{(x - 1)}^{3}} + \frac{D}{{(x - 1)}^{4}},$ then ${(C)}^{A}$