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# Sum of rational roots of $x^5=\large\frac{133x-78}{133-78x}$ is

$\begin{array}{1 1}(A)\;\large\frac{2}{9}&(B)\;\large\frac{9}{2}\\(C)\;\large\frac{6}{13}&(D)\;\large\frac{13}{6}\end{array}$

$x^5=\large\frac{133x-78}{133-78x}$
$\Rightarrow 78x^6-133x^5+133x-78=0$
$x=1,-1$ are roots
Dividing by $(x^2-1)$ we get
$78x^4-133x^3+78x^2-133x+78=0$
Dividing by $x^2$ we get
$78x^2-133x+78-\large\frac{133}{x}+\frac{78}{x^2}$$=0 \Rightarrow 78(x^2+\large\frac{1}{x^2})$$-133(x+\large\frac{1}{x})$$+78=0 Putting x+\large\frac{1}{x}$$=y$ we get
$78y^2-133y-78=0$
$\Rightarrow y=\large\frac{13}{6},-\frac{6}{13}$
$\Rightarrow x+\large\frac{1}{x}=\frac{13}{6}$
$\Rightarrow x=\large\frac{2}{3},\frac{3}{2}$
$x+\large\frac{1}{x}=-\frac{6}{13}$
$\Rightarrow$ Gives imaginary roots
Hence rational roots are $1,-1,\large\frac{2}{3},\frac{3}{2}$
Sum of rational roots =$\large\frac{13}{6}$
Hence (D) is the correct answer.