Question

(c-3)^2/c^2-9 сколько получается?

Answer 1

$\displaystyle \tt\frac{(c-3)^{2}}{c^{2}-9}=\frac{(c-3)(c-3)}{(c-3)(c+3)}=\frac{c-3}{c+3};$

Answer 2

(с-3)*(с-3) \(с-3)*(с+3)=с-3 \с+3

Answer 3

To simplify the given expression, let's expand the numerator and factor the denominator:

Expression: $\frac{(c-3)^2}{c^2-9}$

Step 1: Expand the numerator $(c-3)^2$ $(c-3)^2 = (c-3)(c-3) = c^2 - 6c + 9$

Step 2: Factor the denominator $c^2 - 9$ $c^2 - 9$ is the difference of squares and can be factored as $(c+3)(c-3)$

Now, the expression becomes:

$\frac{c^2 - 6c + 9}{(c+3)(c-3)}$

Step 3: Simplify the expression by canceling out the common factors in the numerator and denominator: The expression has a common factor of $(c-3)$ in both the numerator and denominator, so we can cancel it out:

$\frac{c^2 - 6c + 9}{(c+3)(c-3)} = \frac{(c-3)(c-3)}{(c+3)(c-3)}$

$= \frac{c-3}{c+3}$

Thus, the simplified expression is $\frac{c-3}{c+3}$.