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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) ધારો કે $y = x^2$. પદાવલિ $\frac{y}{(y+2)(y^2-1)} = \frac{y}{(y+2)(y-1)(y+1)} = \frac{A}{y-1} + \frac{B}{y+1} + \frac{C}{y+2}$ બને છે.આંશિક અપૂર્ણ

Vedclass · Accepted Answer

$\frac{4}{3}$

Vedclass · Answer

(B) ધારો કે $y = x^2$. પદાવલિ $\frac{y}{(y+2)(y^2-1)} = \frac{y}{(y+2)(y-1)(y+1)} = \frac{A}{y-1} + \frac{B}{y+1} + \frac{C}{y+2}$ બને છે.આંશિક અપૂર્ણાંક વિઘટનનો ઉપયોગ કરતા:$y = A(y+1)(y+2) + B(y-1)(y+2) + C(y-1)(y+1)$.$y=1$ માટે: $1 = A(2)(3) \implies 6A = 1 \implies A = \frac{1}{6}$.$y=-1$ માટે: $-1 = B(-2)(1) \implies -2B = -1 \implies B = \frac{1}{2}$.$y=-2$ માટે: $-2 = C(-3)(-1) \implies 3C = -2 \implies C = -\frac{2}{3}$.આમ,$A+B-C = \frac{1}{6} + \frac{1}{2} - (-\frac{2}{3}) = \frac{1}{6} + \frac{3}{6} + \frac{4}{6} = \frac{8}{6} = \frac{4}{3}$.

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

Similar Questions

જો $\frac{-x^2+6x+1}{(x-1)^2(x^2+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx-3}{x^2+2}$ હોય,તો $A+B+C=$

જો $\frac{x^3}{(2x - 1)(x + 2)(x - 3)} = p + \frac{q}{2x - 1} + \frac{r}{x + 2} + \frac{s}{x - 3}$ હોય,તો:

જો $\frac{x^2}{(x^2 + a^2)(x^2 + b^2)} = k \left( \frac{a^2}{x^2 + a^2} - \frac{b^2}{x^2 + b^2} \right)$ હોય,તો $k =$

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module