Answer:
option D
[tex]\frac{2}{(k+2)}[/tex]
Explanation:
Given in the question an expression
[tex]\frac{4k+2}{k^{2}-4}.\frac{k-2}{2k+1}[/tex]
Step 1
Use Algebraic Formula
a² - b² = (a-b)(a+b)
k² - 4 = k² - (2)² = (k-2)(k+2)
[tex]\frac{4k+2}{(k-2)(k+2)}.\frac{k-2}{2k+1}[/tex]
Step 2
Cancel(k-2) from both numerator and denometor
[tex]\frac{4k+2}{(k+2)}.\frac{1}{2k+1}[/tex]
Step 3
Use Distributive Law
a(b+c) = (ab + ac)
4k + 2 = 2(2k+1)
[tex]\frac{2(2k+1)}{(k+2)}.\frac{1}{2k+1}[/tex]
Step 4
Cancel(2k+1) from both numerator and denometor
[tex]\frac{2}{(k+2)}.\frac{1}{1}[/tex]
Step 5
Simplified form is
[tex]\frac{2}{(k+2)}[/tex]
