Question

drag expressions to complete the solution.
\\(\left(\frac{x^2 - 4}{2x^2} \cdot \frac{4x}{x^2 - 4x - 12}\
ight) + \frac{2}{x^2 - 6x} = \square + \frac{2}{x^2 - 6x} = \square\\)
\\(\frac{2}{x^2 - 6x}\\) \\(\frac{2x - 2}{x^2 - 6x}\\) \\(\frac{2x - 4}{x^2 - 6x}\\) \\(\frac{4x^3 - 16x + 2}{x^2 - 6x}\\) \\(\frac{4x^3 - 16x}{2x^4 + 8x^2 - 12}\\) \\(\frac{4x^3 + 4x^2 - 8x}{2x^4 - 8x^3 - 24x^2}\\)

Explanation:

Step1: Factor and Simplify the Product

First, factor the expressions: \(x^2 - 4=(x - 2)(x + 2)\), \(x^2-4x - 12=(x - 6)(x + 2)\), and \(x^2-6x=x(x - 6)\). Now, simplify \(\frac{x^2 - 4}{2x^2}\cdot\frac{4x}{x^2 - 4x - 12}\):

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Step2: Add the Two Fractions

Now we have \(\frac{2x - 4}{x^2 - 6x}+\frac{2}{x^2 - 6x}\). Since the denominators are the same, we add the numerators:

$$ \frac{(2x - 4)+2}{x^2 - 6x}=\frac{2x - 2}{x^2 - 6x} $$

Wait, no, let's re - check the first step. Wait, when we multiply \(\frac{x^{2}-4}{2x^{2}}\cdot\frac{4x}{x^{2}-4x - 12}\):

\(x^{2}-4=(x - 2)(x + 2)\), \(x^{2}-4x - 12=(x - 6)(x + 2)\)

So \(\frac{(x - 2)(x + 2)}{2x^{2}}\times\frac{4x}{(x - 6)(x + 2)}=\frac{(x - 2)\times4x}{2x^{2}\times(x - 6)}=\frac{4x(x - 2)}{2x^{2}(x - 6)}=\frac{2(x - 2)}{x(x - 6)}=\frac{2x-4}{x^{2}-6x}\) (because \(x(x - 6)=x^{2}-6x\))

Then add \(\frac{2x - 4}{x^{2}-6x}+\frac{2}{x^{2}-6x}=\frac{2x-4 + 2}{x^{2}-6x}=\frac{2x-2}{x^{2}-6x}\)

Wait, but let's check the numerator of the first product again. Wait, the original first fraction is \(\frac{x^{2}-4}{2x^{2}}\) and the second is \(\frac{4x}{x^{2}-4x - 12}\).

Multiplying numerators: \((x^{2}-4)\times4x=(x - 2)(x + 2)\times4x\)

Multiplying denominators: \(2x^{2}\times(x^{2}-4x - 12)=2x^{2}(x - 6)(x + 2)\)

Cancel out \((x + 2)\) and \(2x\) (from \(4x\) and \(2x^{2}\)):

\(\frac{(x - 2)\times2}{x(x - 6)}=\frac{2x-4}{x^{2}-6x}\) (since \(x(x - 6)=x^{2}-6x\))

Then adding \(\frac{2x - 4}{x^{2}-6x}+\frac{2}{x^{2}-6x}=\frac{2x-4 + 2}{x^{2}-6x}=\frac{2x - 2}{x^{2}-6x}\)

Wait, but let's check the options. The first blank is the result of \(\frac{x^{2}-4}{2x^{2}}\cdot\frac{4x}{x^{2}-4x - 12}\), which we found to be \(\frac{2x - 4}{x^{2}-6x}\). Then the second blank is \(\frac{2x - 4+2}{x^{2}-6x}=\frac{2x - 2}{x^{2}-6x}\)

Answer:

First blank: \(\boldsymbol{\frac{2x - 4}{x^{2}-6x}}\)

Second blank: \(\boldsymbol{\frac{2x - 2}{x^{2}-6x}}\)