If 1-x4 - x2 - 1 - Ax - B-x2 - x - 1 - Cx - D-x2 - x - 1- then C - D -

The correct option is D 0 Ax+B(x^2+1)+x+Cx+D(x^2+1) x =Ax(x^2+1 x)+B(x^2+1 x)+Cx(x^2+1+x)+D(x^2+1+x)(x^2+1)^2 x^2 =x^3(A+C)+x^2(C+D+B A)+x(A B+C+ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Common Roots

If 1/x4 + x...

Question

If $\frac{1}{x^{4} + x^{2} + 1} = \frac{A x + B}{x^{2} + x + 1} + \frac{C x + D}{x^{2} - x + 1}$ , then $C + D =$

-1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

Suggest Corrections

Similar questions

\frac{3 x^{2} + 1}{(x^{2} + 1)^{3}} = \frac{A x + B}{(x^{2} + 1)} + \frac{C x + D}{(x^{2} + 1)^{2}} + \frac{E x + F}{(x^{2} + 1)^{3}}

A + C + E + F =

$I f \frac{d}{d x} [\frac{1 + x^{2} + x^{4}}{1 + x + x^{2}}] = a x + b t h e n (a, b) =$

\frac{2 x^{3} + x^{2} - 5}{x^{4} - 25} = \frac{A x + B}{x^{2} - 5} + \frac{C x + 1}{x^{2} + 5} \Rightarrow (A, B, C) =

\frac{d}{d x} (\frac{1 + x^{2} + x^{4}}{1 + x + x^{2}}) = a x + b

=

\frac{x^{4} + 24 x^{2} + 28}{(x^{2} + 1)^{3}}

= \frac{A x + B}{x^{2} + 1} + \frac{C x + D}{(x^{2} + 1)^{2}} + \frac{E x + F}{(x^{2} + 1)^{3}}

A =

Join BYJU'S Learning Program