If frac{32x^2+186x}{(x^2+1)(x+5)}=frac{37x+1} | vedclass

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(D) Given the equation: $\frac{32x^2+186x}{(x^2+1)(x+5)}=\frac{37x+1}{x^2+1}+\frac{\lambda}{x+5}$ Combining the terms on the right side: $\frac{32x^2+

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$\frac{-5}{2}$

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(D) Given the equation: $\frac{32x^2+186x}{(x^2+1)(x+5)}=\frac{37x+1}{x^2+1}+\frac{\lambda}{x+5}$ Combining the terms on the right side: $\frac{32x^2+186x}{(x^2+1)(x+5)}=\frac{(37x+1)(x+5)+\lambda(x^2+1)}{(x^2+1)(x+5)}$ Equating the numerators: $32x^2+186x = (37x^2 + 185x + x + 5) + \lambda x^2 + \lambda$ $32x^2+186x = (37+\lambda)x^2 + 186x + (5+\lambda)$ Comparing the coefficients of $x^2$ and the constant terms: $37+\lambda = 32 \implies \lambda = -5$ $5+\lambda = 0 \implies \lambda = -5$ Thus,$\frac{\lambda}{2} = \frac{-5}{2}$.

If $\frac{32x^2+186x}{(x^2+1)(x+5)}=\frac{37x+1}{x^2+1}+\frac{\lambda}{x+5}$,then $\frac{\lambda}{2}$ is equal to

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