Simplify the rational expressions. state any excluded values.

1. 2x-8/x-4

2. 4x-8/4x+20

3. x+7/x^2+4x-21

4. x^2-3x-10/x+2

5. x^2-4/2-x

I need help pleeeese

Respuesta :

See the answers in explanation

Explanation:

Let's solve this problem as follows:

First.

[tex]\bullet \ \frac{2x-8}{x-4} \\ \\ Common \ factor \ 2 \ from \ the \ numerator: \\ \\ \frac{2x-8}{x-4}=\frac{2(x-4)}{x-4} =2[/tex]

Second.

[tex]\bullet \ \frac{4x-8}{4x+20} \\ \\ Common \ factor \ 4 \ from \ the \ numerator \ and \ denominator: \\ \\ \frac{4x-8}{4x+20}=\frac{4(x-2)}{4(x+5)}=\frac{(x-2)}{(x+5)}[/tex]

Third

[tex]\bullet \ \frac{x+7}{x^2+4x-21} \\ \\ Rearranging \ denominator: \\ \\ \frac{x+7}{x^2-3x+7x-21}=\frac{x+7}{x(x-3)+7(x-3)}=\frac{x+7}{x(x-3)+7(x-3)} \\ \\ Common \ factor \ x-3 \ from \ denominator: \\ \\ \frac{x+7}{(x-3)(x+7)}=\frac{1}{x-3}[/tex]

Fourth.

[tex]\bullet \ \frac{x^2-3x-10}{x+2} \\ \\ Rearranging \ numerator: \\ \\ \frac{x^2-3x-10}{x+2}=\frac{x^2-5x+2-10}{x+2}=\frac{x(x-5)+2(x-5)}{x+2} \\ \\ Common \ factor \ x-5 \\ \\ \frac{(x-5)(x+2)}{x+2}=x-5[/tex]

Fifth.

[tex]\bullet \ \frac{x^2-4}{2-x} \\ \\ Difference \ of \ squares \ from \ numerator: \\ \\ \frac{(x-2)(x+2)}{2-x} \\ \\ Common \ factor \ -1 \ from \ denominator: \\ \\ \frac{(x-2)(x+2)}{-(x-2)}=-(x+2)[/tex]

Learn more:

Simplification of complex numbers: https://brainly.com/question/2284873

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