Answer :
Answer:
[tex]\frac{4k + 2}{k^{2}-4 }[/tex] × [tex]\frac{k-2}{2k+1}[/tex] = [tex]\frac{2}{k + 2}[/tex]
Step-by-step explanation:
[tex]\frac{4k + 2}{k^{2}-4 }[/tex] × [tex]\frac{k-2}{2k+1}[/tex]
To solve the above, we need to follow the steps below;
4k+2 can be factorize, so that;
4k +2 = 2 (2k + 1)
k² - 4 can also be be expanded, so that;
k² - 4 = (k-2)(k+2)
Lets replace 4k +2 by 2 (2k + 1)
and
k² - 4 by (k-2)(k+2) in the expression given
[tex]\frac{4k + 2}{k^{2}-4 }[/tex] × [tex]\frac{k-2}{2k+1}[/tex]
[tex]\frac{2(2k+ 1)}{(k-2)(k+2)}[/tex] × [tex]\frac{k-2}{2k+1}[/tex]
(2k+1) at the numerator will cancel-out (2k+1) at the denominator, also (k-2) at the numerator will cancel-out (k-2) at the denominator,
So our expression becomes;
[tex]\frac{2}{k + 2}[/tex]
Therefore, [tex]\frac{4k + 2}{k^{2}-4 }[/tex] × [tex]\frac{k-2}{2k+1}[/tex] = [tex]\frac{2}{k + 2}[/tex]