Solve this
Question:

$\frac{1}{(x-2)}+\frac{2}{(x-1)}=\frac{6}{x},(x \neq 2,1)$

Solution:

Given :

$\frac{1}{(x-2)}+\frac{2}{(x-1)}=\frac{6}{x}$

$\Rightarrow \frac{(x-1)+2(x-2)}{(x-1)(x-2)}=\frac{6}{x}$

$\Rightarrow \frac{3 x-5}{x^{2}-3 x+2}=\frac{6}{x}$

$\Rightarrow 3 x^{2}-5 x=6 x^{2}-18 x+12 \quad[$ On cross multiplying $]$

$\Rightarrow 3 x^{2}-13 x+12=0$

$\Rightarrow 3 x^{2}-(9+4) x+12=0$

$\Rightarrow 3 x^{2}-9 x-4 x+12=0$

$\Rightarrow 3 x(x-3)-4(x-3)=0$

$\Rightarrow(3 x-4)(x-3)=0$

$\Rightarrow 3 x-4=0$ or $x-3=0$

$\Rightarrow x=\frac{4}{3}$ or $x=3$

Hence, the roots of the equation are $\frac{4}{3}$ and 3 .