જો frac{2x^3+1}{2x^2-x-6} = ax+b+frac{A}{px-2 | vedclass

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(A) આપેલ પદાવલિ $\frac{2x^3+1}{2x^2-x-6} = ax+b+\frac{A}{px-2}+\frac{B}{2x+q}$ છે.$2x^3+1$ ને $2x^2-x-6$ વડે ભાગતા:$\frac{2x^3+1}{2x^2-x-6} = (x + \fr

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$23$

Vedclass · Answer

(A) આપેલ પદાવલિ $\frac{2x^3+1}{2x^2-x-6} = ax+b+\frac{A}{px-2}+\frac{B}{2x+q}$ છે.$2x^3+1$ ને $2x^2-x-6$ વડે ભાગતા:$\frac{2x^3+1}{2x^2-x-6} = (x + \frac{1}{2}) + \frac{\frac{17}{2}x+4}{2x^2-x-6}$.છેદના અવયવ પાડતા: $2x^2-x-6 = (x-2)(2x+3)$.$\frac{\frac{17}{2}x+4}{(x-2)(2x+3)} = \frac{A}{x-2} + \frac{B}{2x+3}$ માટે આંશિક અપૂર્ણાંકનો ઉપયોગ કરતા.$A = \frac{17}{7}$ અને $B = \frac{23}{14}$ મળે છે.સરખામણી કરતા $a=1, b=\frac{1}{2}, p=1, q=3, A=\frac{17}{7}, B=\frac{23}{14}$ મળે છે.$51apB = 51 	imes 1 	imes 1 	imes \frac{23}{14} = \frac{1173}{14}$.$23bqA = 23 	imes \frac{1}{2} 	imes 3 	imes \frac{17}{7} = \frac{1173}{14}$.આમ,$51apB = 23bqA$.

જો $\frac{2x^3+1}{2x^2-x-6} = ax+b+\frac{A}{px-2}+\frac{B}{2x+q}$ હોય,તો $51apB=$ ($bqA$ માં)

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