What is \frac{x^2+ab}{x^2+m^2ab} equal to ?
Consider the following for the next two (02) items that follow : Let \frac{(x-a)(x-b)}{(x-ma)(x-mb)}=\frac{(x+a)(x+b)}{(x+ma)(x+mb)}; m, a, b \gt 0.
- A. -\frac{1}{m^2}
- B. \frac{1}{m^2}
- C. \frac{2}{m}
- D. \frac{1}{m} ✓
Correct Answer: D. \frac{1}{m}
Explanation
Let S = (a+b)x. The equation is \frac{x^2-S+ab}{x^2-mS+m^2ab} = \frac{x^2+S+ab}{x^2+mS+m^2ab}. Let N = x^2+ab and D = x^2+m^2ab. Then \frac{N-S}{D-mS} = \frac{N+S}{D+mS}. Cross-multiplying yields 2mSN = 2SD, so mN = D, meaning \frac{N}{D} = \frac{1}{m}.
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