29. $$ \frac{x^{2}-4}{x+1} \div \frac{x+2}{x-2} $$

Question

UpStudy AI · Accepted Answer

1. Write the given expression:
$$
\frac{x^2-4}{x+1} \div \frac{x+2}{x-2}
$$

2. Factor the numerator $$x^2-4$$ as a difference of two squares:
$$
x^2-4=(x-2)(x+2)
$$
Thus the expression becomes:
$$
\frac{(x-2)(x+2)}{x+1} \div \frac{x+2}{x-2}
$$

3. Replace the division by multiplication by the reciprocal:
$$
\frac{(x-2)(x+2)}{x+1} \times \frac{x-2}{x+2}
$$

4. Cancel the common factor $$(x+2)$$ (provided that $$x+2\neq0$$, i.e., $$x\neq -2$$):
$$
\frac{(x-2)\cancel{(x+2)}}{x+1} \times \frac{x-2}{\cancel{(x+2)}} = \frac{(x-2)(x-2)}{x+1}
$$

5. Write the final simplified expression:
$$
\frac{(x-2)^2}{x+1}
$$

6. Note the domain restrictions:
- $$x+1\neq 0$$ so $$x\neq -1$$
- $$x-2\neq 0$$ (from the original divisor) so $$x\neq 2$$
- $$x+2\neq 0$$ (for cancellation) so $$x\neq -2$$

Thus, the simplified result is:
$$
\frac{(x-2)^2}{x+1} \quad \text{with } x\neq -1,\; x\neq 2,\; x\neq -2.
$$
The simplified expression is $$ \frac{(x-2)^2}{x+1} $$ with the restrictions $$ x \neq -1 $$, $$ x \neq 2 $$, and $$ x \neq -2 $$.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Extra Insights

preguntas relacionadas

Latest Algebra Questions