જો frac{27x^2+32x+16}{(3x+2)^2(1-x)} = frac{A | vedclass

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(B) આપેલ આંશિક અપૂર્ણાંક વિઘટન: $\frac{27x^2+32x+16}{(3x+2)^2(1-x)} = \frac{A}{3x+2} + \frac{B}{(3x+2)^2} + \frac{C}{1-x}$.બંને બાજુ $(3x+2)^2(1-x)$ વ

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) આપેલ આંશિક અપૂર્ણાંક વિઘટન: $\frac{27x^2+32x+16}{(3x+2)^2(1-x)} = \frac{A}{3x+2} + \frac{B}{(3x+2)^2} + \frac{C}{1-x}$.બંને બાજુ $(3x+2)^2(1-x)$ વડે ગુણતા:$27x^2+32x+16 = A(3x+2)(1-x) + B(1-x) + C(3x+2)^2$.$x = 1$ લેતા: $27(1)^2 + 32(1) + 16 = C(3(1)+2)^2$ $\Rightarrow 75 = 25C$ $\Rightarrow C = 3$.$x = -2/3$ લેતા: $27(-2/3)^2 + 32(-2/3) + 16 = B(1 - (-2/3))$ $\Rightarrow 12 - 64/3 + 16 = B(5/3)$ $\Rightarrow 20/3 = 5B/3$ $\Rightarrow B = 4$.$x = 0$ લેતા: $16 = A(2)(1) + B(1) + C(2)^2$ $\Rightarrow 16 = 2A + 4 + 4(3)$ $\Rightarrow 16 = 2A + 16$ $\Rightarrow A = 0$.અંતે,$AB + BC + CA = (0)(4) + (4)(3) + (3)(0) = 0 + 12 + 0 = 12$.

જો $\frac{27x^2+32x+16}{(3x+2)^2(1-x)} = \frac{A}{3x+2} + \frac{B}{(3x+2)^2} + \frac{C}{1-x}$ હોય,તો $AB+BC+CA =$

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