Mathematics: Pls help I give points. And if you do understand it could you also explain it it a little. Thank you

1. 2. The excluded values are x = 0 (not removable), x = -2 (not removable), and x = 2 (removable)Explanation:The excluded values are those that make any denominator 0.1. Excluded values for 2. Excluded values for 3. Join both setsThe excluded values are x = 0, x = -2, and x = 24. Simplify At the end, the factor x - 2 was canceled, meaning that the excluded value x = 2 is removable.x...

Pls help I give points. And if you do understand

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greg777

07.10.2021

Mathematics

1. $\dfrac{x+1}{x(x+2)}$

2. The excluded values are x = 0 (not removable), x = -2 (not removable), and x = 2 (removable)

Explanation:

The excluded values are those that make any denominator 0.

1. Excluded values for

$-\dfrac{2}{x^2-4}$

$x^2-4=0\implies x^2=4\implies x=\pm2$

2. Excluded values for

$\dfrac{x-1}{x^2-2x}$

$x^2-2x=0\implies x(x-2)\implies x=0\text{ }and\text{ } x=2$

3. Join both sets

The excluded values are x = 0, x = -2, and x = 2

4. Simplify

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

At the end, the factor x - 2 was canceled, meaning that the excluded value x = 2 is removable.x = -2 and x = 0 yet make the final expression undefined, thus they are not removable.

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Pls help I give points. And if you do understand it could you also explain it it a little. Thank you

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