2234\sqrt[4]{\sqrt[3]{2^2}}4322 is equal to
2−162^{-\dfrac{1}{6}}2−61
2-6
2162^{\dfrac{1}{6}}261
26
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Given,
⇒2234=((22)13)14=22×13×14=216.\Rightarrow \sqrt[4]{\sqrt[3]{2^2}} = ((2^2)^{\dfrac{1}{3}})^{\dfrac{1}{4}} \\[1em] = 2^{2 \times \dfrac{1}{3} \times \dfrac{1}{4}} \\[1em] = 2^{\dfrac{1}{6}}.⇒4322=((22)31)41=22×31×41=261.
Hence, Option 3 is the correct option.
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Solve the following equations:
(i) 3(2x + 1) - 2x + 2 + 5 = 0
(ii) 3x = 9.3y, 8.2y = 4x.
The value of (5116)−34\Big(5\dfrac{1}{16}\Big)^{-\dfrac{3}{4}}(5161)−43 is
49\dfrac{4}{9}94
94\dfrac{9}{4}49
278\dfrac{27}{8}827
827\dfrac{8}{27}278
The product 23.24.3212\sqrt[3]{2}.\sqrt[4]{2}.\sqrt[12]{32}32.42.1232 equals
2\sqrt{2}2
2
212\sqrt[12]{2}122
3212\sqrt[12]{32}1232
The value of (81)−24\sqrt[4]{(81)^{-2}}4(81)−2 is
19\dfrac{1}{9}91
13\dfrac{1}{3}31
9
181\dfrac{1}{81}811
