$\dfrac{1}{x-2}-\dfrac{1}{x+2}=\dfrac{x+2}{(x-2)(x+2)}-\dfrac{x-2}{(x-2)(x+2)}\\\\=\dfrac{(x+2)-(x-2)}{(x-2)(x+2)}=\dfrac{x+2-x+2}{(x-2)(x+2)}=\dfrac{4}{x^2-4}\to(D)$

Used: $a^2-b^2=(a-b)(a+b)$

Answer: (D)=(4/x^(2)-4)

5 0

3 years ago

Select all that are equivalent to a^2/a^9 <br> Assume a \= 0<br> 1<br> 1<br> DONE

Verdich [7]

0^2/0^9 = 0/0, therefore your answer is 1/1 considering that it is a fraction that is 100%, but if you divide 0 by 0, according to Keith’s theory you should have an endless number aka Infinity.

6 0

3 years ago

Need help please, please make it simple i just need the answer no explination.

Veronika [31]

Answer:

$\frac{6}{7}$ * $\frac{7}{6}$ = $\frac{42}{42}$ = 1

Step-by-step explanation:

4 0

3 years ago

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