જો frac{x^4+24 x^2+28}{left(x^2+1right)^3}=fr | vedclass

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Question

(C) આપેલ છે $\frac{x^4+24 x^2+28}{\left(x^2+1\right)^3} = \frac{A x+B}{x^2+1} + \frac{C x+D}{\left(x^2+1\right)^2} + \frac{E x+F}{\left(x^2+1\right)^3

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(C) આપેલ છે $\frac{x^4+24 x^2+28}{\left(x^2+1\right)^3} = \frac{A x+B}{x^2+1} + \frac{C x+D}{\left(x^2+1\right)^2} + \frac{E x+F}{\left(x^2+1\right)^3}$.બંને બાજુ $(x^2+1)^3$ વડે ગુણતા:$x^4+24 x^2+28 = (A x+B)(x^2+1)^2 + (C x+D)(x^2+1) + (E x+F)$.જમણી બાજુનું વિસ્તરણ કરતા:$x^4+24 x^2+28 = A x^5 + B x^4 + (2A+C) x^3 + (2B+D) x^2 + (A+C+E) x + (B+D+F)$.સહગુણકોની સરખામણી કરતા:$A = 0, B = 1, C = 0, D = 22, E = 0, F = 5$.તેથી,$A+B+C+D+E+F = 0+1+0+22+0+5 = 28$.

જો $\frac{x^4+24 x^2+28}{\left(x^2+1\right)^3}=\frac{A x+B}{x^2+1}+\frac{C x+D}{\left(x^2+1\right)^2}+\frac{E x+F}{\left(x^2+1\right)^3}$ હોય,તો $A+B+C+D+E+F$ ની કિંમત શોધો.

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