જો frac{x^4+3 x+1}{(x+1)^2(x-1)}=A x+B+frac{C | vedclass

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(D) આપેલ પદાવલિ: $\frac{x^4+3 x+1}{(x+1)^2(x-1)}=A x+B+\frac{C}{x+1}+\frac{D}{(x+1)^2}+\frac{E}{x-1}$.પ્રથમ,$x^4+3x+1$ ને $(x+1)^2(x-1) = x^3+x^2-x-1$

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(D) આપેલ પદાવલિ: $\frac{x^4+3 x+1}{(x+1)^2(x-1)}=A x+B+\frac{C}{x+1}+\frac{D}{(x+1)^2}+\frac{E}{x-1}$.પ્રથમ,$x^4+3x+1$ ને $(x+1)^2(x-1) = x^3+x^2-x-1$ વડે ભાગતા,ભાગફળ $(x-1)$ અને શેષ $3x^2+3x$ મળે છે. તેથી,$Ax+B = x-1$,જેનો અર્થ છે $A=1$ અને $B=-1$.હવે,$\frac{3x^2+3x}{(x+1)^2(x-1)} = \frac{C}{x+1} + \frac{D}{(x+1)^2} + \frac{E}{x-1}$.બંને બાજુ $(x+1)^2(x-1)$ વડે ગુણતા: $3x^2+3x = C(x+1)(x-1) + D(x-1) + E(x+1)^2$.$x=1$ લેતા: $3(1)^2+3(1) = E(1+1)^2 \Rightarrow 6 = 4E \Rightarrow E = 3/2$.$x=-1$ લેતા: $3(-1)^2+3(-1) = D(-1-1) \Rightarrow 0 = -2D \Rightarrow D = 0$.$x^2$ ના સહગુણકો સરખાવતા: $3 = C + E \Rightarrow 3 = C + 3/2 \Rightarrow C = 3/2$.આમ,$A=1, B=-1, C=3/2, D=0, E=3/2$.તેથી,$A+B+C+D+E = 1 - 1 + 3/2 + 0 + 3/2 = 6/2 = 3$.

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

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