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

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(C) આપેલ છે,$\frac{x^3}{(2x - 1)(x - 1)^2} = A + \frac{B}{2x - 1} + \frac{C}{x - 1} + \frac{D}{(x - 1)^2}$.બહુપદી ભાગાકાર કરતા,$\frac{x^3}{2x^3 - 5x^2

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(C) આપેલ છે,$\frac{x^3}{(2x - 1)(x - 1)^2} = A + \frac{B}{2x - 1} + \frac{C}{x - 1} + \frac{D}{(x - 1)^2}$.બહુપદી ભાગાકાર કરતા,$\frac{x^3}{2x^3 - 5x^2 + 4x - 1} = \frac{1}{2} + \frac{5x^2 - 4x + 1}{2(2x - 1)(x - 1)^2}$.અહીં $A = \frac{1}{2}$ મળે છે.હવે,$\frac{5x^2 - 4x + 1}{2(2x - 1)(x - 1)^2} = \frac{B}{2x - 1} + \frac{C}{x - 1} + \frac{D}{(x - 1)^2}$ લેતા.$x = 1$ મુકતા,$D = 1$ મળે છે.$x = \frac{1}{2}$ મુકતા,$B = \frac{1}{2}$ મળે છે.$x^2$ ના સહગુણકોની સરખામણી કરતા,$C = 1$ મળે છે.તેથી,$2A - 3B + 4C + 5D = 2(\frac{1}{2}) - 3(\frac{1}{2}) + 4(1) + 5(1) = 1 - 1.5 + 9 = 8.5 = \frac{17}{2}$.

જો $\frac{x^3}{(2x - 1)(x - 1)^2} = A + \frac{B}{2x - 1} + \frac{C}{x - 1} + \frac{D}{(x - 1)^2}$ હોય,તો $2A - 3B + 4C + 5D = $

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