Question

\(\left(\dfrac{x^{m^2}}{x^{n^2}}\right)^{\frac{1{m-n}}=\)}

• A. \(x^{m-n}\)
• B. \(x^m\)
• C. \(x^{m+n}\)
• D. \(x^n\)

Hint:

Whenever you see \(m^2-n^2\), immediately factor it as \((m-n)(m+n)\). It is one of the most frequently used algebraic identities.

Answer 1

The Correct Option is C

Concept: Use the laws of exponents: \[ \frac{x^a}{x^b}=x^{a-b} \] and \[ (x^a)^b=x^{ab} \]

Step 1: Simplifying the fraction.
\[ \left(\frac{x^{m^2}}{x^{n^2}}\right)^{\frac1{m-n}} = \left(x^{m^2-n^2}\right)^{\frac1{m-n}} \]

Step 2: Using difference of squares.
\[ m^2-n^2=(m-n)(m+n) \] Therefore, \[ \left(x^{(m-n)(m+n)}\right)^{\frac1{m-n}} \]

Step 3: Applying the power rule.
\[ x^{\frac{(m-n)(m+n)}{m-n}} \] \[ =x^{m+n} \] Hence, \[ \boxed{x^{m+n}} \] {\(x^{m+n}\)}