128×32(−43)128 \times 32^{\Big(-\dfrac{4}{3}\Big)}128×32(−34) =
43\sqrt[3]{4}34
23\sqrt[3]{2}32
8
2
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Given,
⇒128×32(−43)⇒(2)7×[(2)5]−43⇒(2)7×(2)5×−43⇒(2)7×(2)−203⇒(2)7−203⇒(2)21−203⇒(2)13⇒23.\Rightarrow 128 \times 32^{\Big(\dfrac{-4}{3}\Big)} \\[1em] \Rightarrow (2)^7 \times [(2)^5]^{\dfrac{-4}{3}} \\[1em] \Rightarrow (2)^7 \times (2)^{\dfrac{5 \times -4}{3}} \\[1em] \Rightarrow (2)^7 \times (2)^{\dfrac{-20}{3}} \\[1em] \Rightarrow (2)^{7 - \dfrac{20}{3}} \\[1em] \Rightarrow (2)^{\dfrac{21 - 20}{3}} \\[1em] \Rightarrow (2)^{\dfrac{1}{3}} \\[1em] \Rightarrow \sqrt[3]{2}.⇒128×32(3−4)⇒(2)7×[(2)5]3−4⇒(2)7×(2)35×−4⇒(2)7×(2)3−20⇒(2)7−320⇒(2)321−20⇒(2)31⇒32.
Hence, option 2 is the correct option.
Answered By
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[(a43)−32]−12\Big[\Big(\sqrt[3]{a^4}\Big)^{\dfrac{-3}{2}}\Big]^{\dfrac{-1}{2}}[(3a4)2−3]2−1 =
a
a2
1a\dfrac{1}{a}a1
1a2\dfrac{1}{a^2}a21
5×3−3×5×36656×3\dfrac{\sqrt{5 \times 3^{-3}} \times \sqrt[6]{5 \times 3^6}}{\sqrt[6]{5} \times \sqrt{3}}65×35×3−3×65×36 =
35\sqrt{\dfrac{3}{5}}53
53\sqrt{\dfrac{5}{3}}35
35\dfrac{3}{5}53
53\dfrac{\sqrt{5}}{3}35
