Question

(1/3)² ; (-3/7)² ; (9/10)⁰

Answer 1

\(\left(\dfrac{1}{3}\right)^2=\dfrac{1}{3}\cdot\dfrac{1}{3}=\dfrac{1}{9}\)

\(\left(-\dfrac{3}{7}\right)^2=\left(-\dfrac{3}{7}\right)\cdot\left(-\dfrac{3}{7}\right)=\dfrac{9}{49}\)

\(\left(\dfrac{9}{10}\right)^0=1\)

Answer 2

\(\left(\dfrac{1}{3}\right)^2\)=  \(\dfrac{\left(1\right)^2}{\left(3\right)^2}\) = \(\dfrac{1}{9}\) .

\(\left(-\dfrac{3}{7}\right)^2\) = \(\dfrac{\left(-3\right)^2}{\left(7\right)^2}\) = \(\dfrac{9}{49}.\)

\(\left(\dfrac{9}{10}\right)^0=\dfrac{\left(9\right)^0}{\left(10\right)^0}=1\) . (Vì \(a^0=1\) nên \(\left(\dfrac{9}{10}\right)^0=1\))