Question

can somebody help me? fan and medal.

Answer 1

Okay, so you have to know that when you multiply

\(\LARGE\color{black}{ \frac{B}{A} \times \frac{C}{B} }\) then the Bs cancel, and you get C/A.
(right?)

Then, you also have to know that,
\(\normalsize\color{blue}{ a^2-b^2=(a-b)(a+b)}\)

Answer 2

\(\bf \cfrac{\cancel{ (x+3) }}{(x-3)}\times \cfrac{(x^2-9)}{\cancel{(x+3) }}\qquad
\begin{cases}
{\color{blue}{ a^2-b^2 = (a-b)(a+b)}}
\\ \quad \\
\bf x^2-9\to x^2-3^2\to (\square -\square )(\square +\square )
\end{cases}
\\ \quad \\
\cfrac{(\square -\square )(\square +\square )}{(x-3)}?\)

Answer 3

yess

Answer 4

In your first problem:
1) Is there anything that can be canceled out right away?
2) Can you re-write x^2-9 differently?
If you can what will cancel?

Answer 5

you can cancel out x+3 and you can write that out as (x-9) + (x-9)?