Partial fraction of $\frac{5 x^{2} - 4}{x^{2} \left ( x - 2 \right )} i s -$  

$\frac{5 x^{2}-4}{x^{2}(x-2)} \equiv \frac{A}{x^{2}}+\frac{B}{x}+\frac{C}{x-2}$

$\Rightarrow 5 \mathrm{x}^{2}-4 \equiv \mathrm{A}(\mathrm{x}-2)+\mathrm{Bx}(\mathrm{x}-2)+\mathrm{Cx}^{2}$

Putting $\mathrm{x}=0$ we get, $-4=-2 \mathrm{~A} \therefore \mathrm{A}=2$

Putting $\mathrm{x}=2$ we get, $16=4 \mathrm{C} \therefore \mathrm{C}=4$

(2) $\Rightarrow 5 \mathrm{x}^{2}-4=2(\mathrm{x}-2)+\mathrm{Bx}(\mathrm{x}-2)+4 \mathrm{x}^{2}$

Putting $\mathrm{x}=1$ we get, $1=-2-\mathrm{B}+4 \therefore \mathrm{B}=1 \therefore(1) \Rightarrow \frac{5 \mathrm{x}^{2}-4}{\mathrm{x}^{2}(\mathrm{x}-2)}=\frac{2}{\mathrm{x}^{2}}+\frac{1}{\mathrm{x}}+\frac{4}{\mathrm{x}-2}=\frac{1}{\mathrm{x}}+\frac{2}{\mathrm{x}^{2}}+\frac{4}{\mathrm{x}-2}$