Question

i want. the answerfast pls ​

Answer 1

Evaluate

$\sf(i)\quad\bigg\{ \bigg(\dfrac{4}{3} \bigg)^{ - 1} -\bigg(\dfrac{1}{4} \bigg)^{ - 1} \bigg\}^{ - 1}\\\\\\:\implies\sf \bigg\{\dfrac{3}{4}-4 \bigg\}^{ - 1}\\\\\\:\implies\sf \bigg\{\dfrac{3 - (4 \times 4)}{4}\bigg\}^{ - 1}\\\\\\:\implies\sf \bigg\{\dfrac{3 - 16}{4}\bigg\}^{ - 1}\\\\\\:\implies\sf \bigg\{\dfrac{- \:13}{4}\bigg\}^{ - 1}\\\\\\:\implies\sf \dfrac{ - \:4}{13}$

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$\sf(ii)\quad\Bigg[\bigg\{\bigg(\dfrac{- \:1}{3}\bigg)^2\bigg\}^{ - 2}\Bigg]^{-1}\\\\\\:\implies\sf \Bigg[\bigg\{\dfrac{1}{9}\bigg\}^{ - 2}\Bigg]^{-1}\\\\\\:\implies\sf \Bigg[\bigg\{9\bigg\}^{2}\Bigg]^{-1}\\\\\\:\implies\sf \Bigg[81\Bigg]^{-1}\\\\\\:\implies\sf \dfrac{1}{81}$

$\rule{200}{1}$

$\boxed{\begin{minipage}{5 cm}\bf{\dag}\:\:\underline{\text{Law of Exponents :}}\\\\\bigstar\:\:\sf\dfrac{a^m}{a^n} = a^{m - n}\\\\\bigstar\:\:\sf{(a^m)^n = a^{mn}}\\\\\bigstar\:\:\sf(a^m)(a^n) = a^{m + n}\\\\\bigstar\:\:\sf\dfrac{1}{a^n} = a^{-n}\\\\\bigstar\:\:\sf\sqrt[\sf n]{\sf a} = (a)^{\dfrac{1}{n}}\end{minipage}}$