Question

Find the sum.
5k/k^2-2k+1

+

2/k^2+k-2

Answer 1

`\frac{5k}{\underbrace{k^2-2k+1}_((a-b)^2=a^2-2ab+b^2)}+(2)/(k^2+k-2)=(5k)/((k-1)^2)+(2)/(k^2+2k-k-2)\\\\=(5k)/((k-1)^2)+(2)/(k(k+2)-1(k+2))=(5k)/((k-1)^2)+(2)/((k+2)(k-1))\\\\=(5k(k+2))/((k-1)^2(k+2))+(2(k-1))/((k-1)^2(k+2))=(5k^2+10k+2k-2)/((k-1)^2(k+2))`

`=(5k^2+12k-2)/((k-1)^2(k+2))=(5k^2+12k-2)/((k^2-2k+1)(k+2))\\\\=(5k^2+12k-2)/(k^3+2k^2-2k^2-4k+k+2)=(5k^2+12k-2)/(k^3-3k+2)`