If ab=23÷(−23)0;then(ab)−2\dfrac{a}{b} = \dfrac{2}{3} ÷ \Big(-\dfrac{2}{3}\Big)^0; \text{then} \Big(\dfrac{a}{b}\Big)^{-2}ba=32÷(−32)0;then(ba)−2 is equal to:
49\dfrac{4}{9}94
−49-\dfrac{4}{9}−94
94\dfrac{9}{4}49
−94-\dfrac{9}{4}−49
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Given, ab=23÷(−23)0\dfrac{a}{b} = \dfrac{2}{3} ÷ \Big(-\dfrac{2}{3}\Big)^0ba=32÷(−32)0
As we know that a0 = 1
⇒ab=23÷1⇒ab=23⇒(ab)−2=(23)−2⇒(ab)−2=(32)2⇒(ab)−2=94.\Rightarrow \dfrac{a}{b} = \dfrac{2}{3} ÷ 1 \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{2}{3}\\[1em] \Rightarrow \Big(\dfrac{a}{b}\Big)^{-2} = \Big(\dfrac{2}{3}\Big)^{-2}\\[1em] \Rightarrow \Big(\dfrac{a}{b}\Big)^{-2} = \Big(\dfrac{3}{2}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{a}{b}\Big)^{-2} = \dfrac{9}{4}.⇒ba=32÷1⇒ba=32⇒(ba)−2=(32)−2⇒(ba)−2=(23)2⇒(ba)−2=49.
Hence, option 3 is the correct option.
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32\dfrac{3}{2}23
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