The sum of all integral values of k ( k ≠ 0 ) for which the equation 2 x - 1 - 1 x - 2 = 2 k in x has…

We have,

$\frac{2}{x-1}-\frac{1}{x-2}=\frac{2}{k}$

$\Rightarrow \frac{2x-4-x+1}{(x-1)(x-2)}=\frac{2}{k}$

$\Rightarrow \frac{x-3}{x^2-3x+2}=\frac{2}{k}$

$\Rightarrow kx-3k=2x^2-6x+4$

$\Rightarrow 2x^2-(6+k)x+3k+4=0$

 For no real roots $D<0$ 

$\Rightarrow (6+k{)}^2-8(3k+4)<0$

$\Rightarrow k^2+12k+36-24k-32<0$

$\Rightarrow (k-6{)}^2-32<0$

$\Rightarrow \left|k-6\right|<\sqrt{32}$

$\Rightarrow 6-\sqrt{32}<k<6+\sqrt{32}$

Integral value of $k=1,2,3,4, 5,6,7,8,9,10,11$

Sum$=\frac{11\times 12}{2}=66$