is it right?it it is so how do i need to continue from here?

Sabi has told me that the answer is different from the book answer.    

$$\frac{9}{5a^2-11ab+2b^2}+\frac{17}{10a^2+13ab-3b^2}\\\\
=\frac{9}{5a^2-10ab-1ab+2b^2}+\frac{17}{10a^2-2ab+15ab-3b^2}\\\\
=\frac{9}{5a(a-2b)-b(a-2b}+\frac{17}{2a(5a-b)+3b(5a-b)}\\\\
=\frac{9}{(5a-b)(a-2b)}+\frac{17}{(2a+3b)(5a-b)}\\\\
=\frac{9(2a+3b)}{(5a-b)(a-2b)(2a+3b)}+\frac{17(a-2b)}{(2a+3b)(5a-b)(a-2b)}\\\\
=\frac{18a+27b}{(5a-b)(a-2b)(2a+3b)}+\frac{17a-34b}{(2a+3b)(5a-b)(a-2b)}\\\\
=\frac{18a+27b+17a-34b}{(5a-b)(a-2b)(2a+3b)}\\\\
=\frac{35a-7b}{(5a-b)(a-2b)(2a+3b)}\\\\
=\frac{7(5a-b)}{(5a-b)(a-2b)(2a+3b)}\\\\
=\frac{7}{(a-2b)(2a+3b)}\\\\$$

Is it the same now Sabi ?

Yes, I think it looks good so far, now you just have to get them both with a common denominator.

That will be  (5a-b)(a-2b)(2a+3b)

then you can add the numerators  

yes thank you very much  my answer was wrong because i didnt open the brachets