Solve for x :

$(81)^{\dfrac{3}{4}} - \Big(\dfrac{1}{32}\Big)^{-\dfrac{2}{5}} + x\Big(\dfrac{1}{2}\Big)^{-1}.2^0 = 27$

Given,

$\Rightarrow (81)^{\dfrac{3}{4}} - \Big(\dfrac{1}{32}\Big)^{-\dfrac{2}{5}} + x\Big(\dfrac{1}{2}\Big)^{-1}.2^0 = 27 \\[1em] \Rightarrow (3^4)^{\dfrac{3}{4}} - \Big(\dfrac{1}{2^5}\Big)^{-\dfrac{2}{5}} + (2^{-1})^{-1}x.1 = 27 \\[1em] \Rightarrow 3^{4 \times \dfrac{3}{4}} - (2^{-5})^{-\dfrac{2}{5}} + 2^{-1\times -1}.x = 27 \\[1em] \Rightarrow 3^3 - 2^{-5 \times -\dfrac{2}{5}} + 2x = 27 \\[1em] \Rightarrow 27 - 2^2 + 2x = 27 \\[1em] \Rightarrow 27 - 4 + 2x = 27 \\[1em] \Rightarrow 23 + 2x = 27 \\[1em] \Rightarrow 2x = 27 - 23 \\[1em] \Rightarrow 2x = 4 \\[1em] \Rightarrow x = \dfrac{4}{2} = 2.$

Hence, x = 2.