જો frac{x^4}{(x-1)(x-2)}=f(x)+frac{A}{x-1}+fr | vedclass

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(D) આપેલ છે $\frac{x^4}{(x-1)(x-2)} = f(x) + \frac{A}{x-1} + \frac{B}{x-2}$.પ્રથમ,$\frac{x^4}{x^2-3x+2}$ માટે બહુપદી ભાગાકાર કરો.$x^4 = (x^2-3x+2)(x^2

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$20$

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(D) આપેલ છે $\frac{x^4}{(x-1)(x-2)} = f(x) + \frac{A}{x-1} + \frac{B}{x-2}$.પ્રથમ,$\frac{x^4}{x^2-3x+2}$ માટે બહુપદી ભાગાકાર કરો.$x^4 = (x^2-3x+2)(x^2+3x+7) + (15x-14)$.તેથી,$\frac{x^4}{(x-1)(x-2)} = x^2+3x+7 + \frac{15x-14}{(x-1)(x-2)}$.$\frac{15x-14}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$ માટે આંશિક અપૂર્ણાંકનો ઉપયોગ કરો.$15x-14 = A(x-2) + B(x-1)$.$x=1$ માટે,$15-14 = A(1-2) \implies A = -1$.$x=2$ માટે,$30-14 = B(2-1) \implies B = 16$.આમ,$f(x) = x^2+3x+7$.$f(-2) = (-2)^2 + 3(-2) + 7 = 4 - 6 + 7 = 5$.તેથી,$f(-2)+A+B = 5 + (-1) + 16 = 20$.

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

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