`(49a^(2))/(25b^(2))-(x^(2))/(100y^(2))`= _______

Options

A. `((7a)/(5b)+x/(10y))((7a)/(5b)-x/y)`
B. `((7a)/(5a)+x/(10y))((7a)/(5b)+x/y)`
C. `((7a)/(5b)+x/(10y))((7a)/(5b)-x/(10y))`
D. `((7a)/(5b)+x/(10y))((7a)/(5b)-x/(10y))`

Answer

To solve the expression \(\frac{49a^2}{25b^2} - \frac{x^2}{100y^2}\), we can factor it step by step.

### Step 1: Rewrite the expression
We start with the expression:
\[ \frac{49a^2}{25b^2} - \frac{x^2}{100y^2} \]

### Step 2: Identify perfect squares
Notice that:
- \(49a^2\) is a perfect square, as \(49 = (7)^2\) and \(a^2 = (a)^2\).
- \(25b^2\) is also a perfect square, as \(25 = (5)^2\) and \(b^2 = (b)^2\).
- \(x^2\) is a perfect square, as \(x^2 = (x)^2\).
- \(100y^2\) is a perfect square, as \(100 = (10)^2\) and \(y^2 = (y)^2\).

Thus, we can rewrite the expression as:
\[ \left(\frac{7a}{5b}\right)^2 - \left(\frac{x}{10y}\right)^2 \]

### Step 3: Apply the difference of squares formula
We can use the difference of squares identity, which states that \(A^2 - B^2 = (A - B)(A + B)\). Here, let:
- \(A = \frac{7a}{5b}\)
- \(B = \frac{x}{10y}\)

So we can write:
\[ \left(\frac{7a}{5b}\right)^2 - \left(\frac{x}{10y}\right)^2 = \left(\frac{7a}{5b} - \frac{x}{10y}\right)\left(\frac{7a}{5b} + \frac{x}{10y}\right) \]

### Step 4: Write the final factored form
Thus, the expression can be factored as:
\[ \left(\frac{7a}{5b} - \frac{x}{10y}\right)\left(\frac{7a}{5b} + \frac{x}{10y}\right) \]

### Final Answer
The final factored form of the expression \(\frac{49a^2}{25b^2} - \frac{x^2}{100y^2}\) is:
\[ \left(\frac{7a}{5b} - \frac{x}{10y}\right)\left(\frac{7a}{5b} + \frac{x}{10y}\right) \]

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