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- \(\dfrac{xxxxx}{xx}= x^3\)
- \(\dfrac{xxxy}{xxxxx}=\dfrac y{x^2}\)
- \(\dfrac{5xxx}{25xyy}=\dfrac{x^2}{5y^2}\)
- When you divide expressions with the same base, you subtract the exponents.
- Proof: \(\dfrac{a^n}{a^m}=\dfrac{(a\cdot a\cdot a\cdot …) \ _{n \ times}}{(a\cdot a\cdot a\cdot …) \ _{m \ times}}\) which means that there are \(n \ a\)’s in the numerator and \(m \ a\)’s in the denominator; thus \(\dfrac{a^n}{a^m}=a^{n-m}\) . Recall that when the result is a negative exponent, that \(a^{-1}=\dfrac1a\).
- One method: \(\Bigg(\dfrac{3}{6}\Bigg)\cdot x^{5-3} =\dfrac {x^2}{2}\)
- One method: \(\dfrac{x^{-2}}{x^3}=x^{-2-3}=x^{-5}=\dfrac{1}{x^5}\)
- \(\dfrac{x^2}{x^2}=x^{2-2}=x^0=1\)
