Question

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

Answer 1

$\dfrac{5k}{\underbrace{k^2-2k+1}_{(a-b)^2=a^2-2ab+b^2}}+\dfrac{2}{k^2+k-2}=\dfrac{5k}{(k-1)^2}+\dfrac{2}{k^2+2k-k-2}\\\\=\dfrac{5k}{(k-1)^2}+\dfrac{2}{k(k+2)-1(k+2)}=\dfrac{5k}{(k-1)^2}+\dfrac{2}{(k+2)(k-1)}\\\\=\dfrac{5k(k+2)}{(k-1)^2(k+2)}+\dfrac{2(k-1)}{(k-1)^2(k+2)}=\dfrac{5k^2+10k+2k-2}{(k-1)^2(k+2)}$
$=\dfrac{5k^2+12k-2}{(k-1)^2(k+2)}=\dfrac{5k^2+12k-2}{(k^2-2k+1)(k+2)}\\\\=\dfrac{5k^2+12k-2}{k^3+2k^2-2k^2-4k+k+2}=\dfrac{5k^2+12k-2}{k^3-3k+2}$