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

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(D) We decompose the expression using partial fractions:$\frac{x+2}{(x^2+3)(x^2)(x^2+1)(x^2+2)} = \frac{A'}{x} + \frac{B'}{x^2} + \frac{C'x+D'}{x^2+1}

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$\frac{1}{4}$

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(D) We decompose the expression using partial fractions:$\frac{x+2}{(x^2+3)(x^2)(x^2+1)(x^2+2)} = \frac{A'}{x} + \frac{B'}{x^2} + \frac{C'x+D'}{x^2+1} + \frac{E'x+F'}{x^2+2} + \frac{G'x+H'}{x^2+3}$.By equating coefficients and solving the system of equations,we obtain:$A' = \frac{1}{6}, B' = \frac{1}{3}, C' = -\frac{1}{2}, D' = -1, E' = \frac{1}{2}, F' = 1, G' = -\frac{1}{6}, H' = -\frac{1}{3}$.Substituting these back,we group the terms to match the form given in the question:$\frac{x+2}{(x^2+3)(x^4+x^2)(x^2+2)} = \frac{-\frac{1}{6}x - \frac{1}{3}}{x^2+3} + \frac{\frac{1}{2}x + 1}{x^2+2} + \frac{-\frac{1}{3}x^3 - \frac{2}{3}x^2 + \frac{1}{6}x + \frac{1}{3}}{x^4+x^2}$.Comparing this with the given expression,we identify:$A = -\frac{1}{6}, B = -\frac{1}{3}, C = \frac{1}{2}, D = 1, E = -\frac{1}{3}, F = -\frac{2}{3}, G = \frac{1}{6}, H = \frac{1}{3}$.Finally,calculating the required value:$(E+F)(C+D)(A) = (-\frac{1}{3} - \frac{2}{3})(\frac{1}{2} + 1)(-\frac{1}{6}) = (-1)(\frac{3}{2})(-\frac{1}{6}) = \frac{3}{12} = \frac{1}{4}$.

If $\frac{x+2}{(x^2+3)(x^4+x^2)(x^2+2)} = \frac{Ax+B}{x^2+3} + \frac{Cx+D}{x^2+2} + \frac{Ex^3+Fx^2+Gx+H}{x^4+x^2}$,then find the value of $(E+F)(C+D)(A)$.

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