Answer:
Therefore, the answer is 3
Step-by-step explanation:
The given expression is:
[tex]\frac{3k}{k-2} + \frac{6}{2-k} \\\\Taking\ LCM\\= \frac{3k(2-k)+6(k-2)}{(k-2)(2-k)} \\=\frac{6k-3k^2+6k-12}{(k-2)(2-k)}\\= \frac{-3k^2+12k-12}{(k-2)(2-k)}\\\\Applying\ formula\\=\frac{-3(k^2-4k+4)}{(k-2)(2-k)}\\=\frac{-3((k)^2-2*2*k+(2)^2)}{(k-2)(2-k)}\\=\frac{-3(k-2)^2}{(k-2)(2-k)}\\=\frac{-3(k-2)}{(2-k)}\\=\frac{3(2-k)}{2-k}\\=3[/tex]
