If frac{ax - 1}{(1 - x + x^2)(2 + x)} = frac{ | vedclass

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

Question

(B) Given the equation: $\frac{ax - 1}{(1 - x + x^2)(2 + x)} = \frac{x}{1 - x + x^2} - \frac{1}{2 + x}$ Multiply both sides by the common denominator

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Given the equation: $\frac{ax - 1}{(1 - x + x^2)(2 + x)} = \frac{x}{1 - x + x^2} - \frac{1}{2 + x}$ Multiply both sides by the common denominator $(1 - x + x^2)(2 + x)$: $ax - 1 = x(2 + x) - 1(1 - x + x^2)$ Expand the right side: $ax - 1 = 2x + x^2 - 1 + x - x^2$ Simplify the expression: $ax - 1 = 3x - 1$ Comparing the coefficients of $x$ on both sides,we get: $a = 3$

If $\frac{ax - 1}{(1 - x + x^2)(2 + x)} = \frac{x}{1 - x + x^2} - \frac{1}{2 + x}$,then $a = $

Similar Questions

If $\frac{x}{(1+x^2)(3-2x)} = \frac{Bx+C}{1+x^2} + \frac{A}{3-2x}$,then $C$ is

If $\frac{1}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$,then $(A, B, C) = $

If $\frac{x^2}{2 x^4+7 x^2+6}=\frac{A x+B}{x^2+a}+\frac{C x+D}{a x^2+3}$, then find the value of $A+B+C-2 D$. (in $a$)

The coefficient of $x^5$ in the expansion of $\frac{x^2 + 1}{(x^2 + 4)(x - 2)}$ is

If $\frac{x^4}{(x-1)^2(x+1)}=Ax+B+\frac{P}{(x-1)}+\frac{Q}{(x-1)^2}+\frac{R}{x+1}$,then $2AP-BQ+R=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module