Simplify the following:
1634+2(12)−1(3)016^{\dfrac{3}{4}} + 2\Big(\dfrac{1}{2}\Big)^{-1}(3)^01643+2(21)−1(3)0
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Given,
⇒1634+2(12)−1(3)0=(24)34+2.(2)1.1=23+4=8+4=12.\Rightarrow 16^{\dfrac{3}{4}} + 2\Big(\dfrac{1}{2}\Big)^{-1}(3)^0 = (2^4)^{\dfrac{3}{4}} + 2.(2)^1 .1 \\[1em] = 2^3 + 4 = 8 + 4 = 12.⇒1643+2(21)−1(3)0=(24)43+2.(2)1.1=23+4=8+4=12.
Hence, 1634+2(12)−1(3)016^{\dfrac{3}{4}} + 2\Big(\dfrac{1}{2}\Big)^{-1}(3)^01643+2(21)−1(3)0 = 12.
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