Question

va roggggg urgent....​

Answer 1

$\boxed {a^{m} \cdot a^{n} = a^{m + n} \ \ \ ; \ \ \Big(a^{m}\Big)^{n} = a^{m \cdot n}}$

$\bigg(\dfrac{1}{2} \bigg)^{0} = 1$

$\bigg[\bigg(\dfrac{9}{8}\bigg)^{2}\bigg]^{3} = \bigg(\dfrac{3^{2} }{2^{3} }\bigg)^{2 \cdot 3} = \bigg(\dfrac{3^{2} }{2^{3} }\bigg)^{6} = \dfrac{3^{2 \cdot 6} }{2^{3 \cdot 6} } = \dfrac{3^{12} }{2^{18} }$

$\bigg(\dfrac{2}{3}\bigg)^{12} = \dfrac{2^{12}}{3^{12}}$

⇒

$\bigg(\dfrac{1}{2} \bigg)^{0} - \bigg[\bigg(\dfrac{9}{8}\bigg)^{2}\bigg]^{3} \cdot \bigg(\dfrac{2}{3}\bigg)^{12} = 1 - \dfrac{3^{12} }{2^{18} } \cdot \dfrac{2^{12}}{3^{12}} = 1 - \dfrac{2^{12}}{2^{18}} =$

$= 1 - \dfrac{1}{2^{18-12}} = 1 - \dfrac{1}{2^{6}} = 1 - \dfrac{1}{64} = \dfrac{64-1}{64} = \bf\dfrac{63}{64}$