[(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
1 Like
Given,
[(a43)−32]−12\Big[\Big(\sqrt[3]{a^4}\Big)^{\dfrac{-3}{2}}\Big]^{\dfrac{-1}{2}}[(3a4)2−3]2−1
Simplifying the expression:
⇒[(a4)13]−32×−12⇒[(a4)13]34⇒a4×13×34⇒a1⇒a.\Rightarrow \Big[\Big(a^4\Big)^{\dfrac{1}{3}}\Big]^{\dfrac{-3}{2} \times \dfrac{-1}{2}} \\[1em] \Rightarrow \Big[\Big(a^4\Big)^{\dfrac{1}{3}}\Big]^{\dfrac{3}{4}} \\[1em] \Rightarrow a^{4 \times \dfrac{1}{3} \times \dfrac{3}{4}} \\[1em] \Rightarrow a^1 \\[1em] \Rightarrow a.⇒[(a4)31]2−3×2−1⇒[(a4)31]43⇒a4×31×43⇒a1⇒a.
Hence, option 1 is the correct option.
Answered By
If 2x = 3y = 6-z, show that: 1x+1y+1z=0\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0x1+y1+z1=0
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
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
(81)0.13 × (81)0.12 =
1
3
3\sqrt{3}3
13\dfrac{1}{\sqrt{3}}31
