Answer:
Sum of [tex]\frac{7x}{x^2-4}[/tex] and [tex]\frac{2}{x+2}[/tex] is [tex]\mathbf{\frac{9x-4}{x^2-4}}[/tex]
Option B is correct answer.
Step-by-step explanation:
We need to find sum of [tex]\frac{7x}{x^2-4}[/tex] and [tex]\frac{2}{x+2}[/tex]
Finding sum of [tex]\frac{7}{x^2-4}[/tex] and [tex]\frac{2}{x+2}[/tex]:
[tex]\frac{7x}{x^2-4}+\frac{2}{x+2}[/tex]
We know that [tex]x^2-4 =(x+2)(x-2)[/tex]
Replacing x^2-4
[tex]\frac{7x}{(x+2)(x-2)}+\frac{2}{x+2}[/tex]
Now, taking LCM of (x+2)(x-2) and (x+2) we get (x+2)(x-2)
[tex]=\frac{7x+2(x-2)}{(x+2)(x-2)}\\=\frac{7x+2x-4}{(x+2)(x-2)}\\=\frac{9x-4}{(x+2)(x-2)}\\=\frac{9x-4}{x^2-4}[/tex]
So, Sum of [tex]\frac{7x}{x^2-4}[/tex] and [tex]\frac{2}{x+2}[/tex] is [tex]\mathbf{\frac{9x-4}{x^2-4}}[/tex]
Option B is correct answer.
