જો frac{9x-7}{(x+3)(x^2+1)} = frac{A}{x+3} + | vedclass

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(B) આપેલ છે, $\frac{9x-7}{(x+3)(x^2+1)} = \frac{A}{x+3} + \frac{Bx+C}{x^2+1}$ અંશને સરખાવતા: $9x-7 = A(x^2+1) + (Bx+C)(x+3)$ $9x-7 = (A+B)x^2 + (3B+C)

Vedclass · Accepted Answer

$\frac{-6}{5}$

Vedclass · Answer

(B) આપેલ છે, $\frac{9x-7}{(x+3)(x^2+1)} = \frac{A}{x+3} + \frac{Bx+C}{x^2+1}$ અંશને સરખાવતા: $9x-7 = A(x^2+1) + (Bx+C)(x+3)$ $9x-7 = (A+B)x^2 + (3B+C)x + (A+3C)$ બંને બાજુ સહગુણકોની સરખામણી કરતા: $A+B = 0$ $3B+C = 9$ $A+3C = -7$ સમીકરણો ઉકેલતા, $A = -\frac{17}{5}, B = \frac{17}{5}, C = -\frac{6}{5}$ તેથી, $A+B+C = -\frac{17}{5} + \frac{17}{5} - \frac{6}{5} = -\frac{6}{5}$

જો $\frac{9x-7}{(x+3)(x^2+1)} = \frac{A}{x+3} + \frac{Bx+C}{x^2+1}$, જ્યાં $A, B, C \in \mathbb{R}$, તો $A+B+C = $

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