\(a,x+\dfrac{1}{2}=\dfrac{6}{4}\)
\(\Rightarrow x=\dfrac{6}{4}-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{6}{4}-\dfrac{2}{4}\)
\(\Rightarrow x=1\)
Vậy `x = 1`
\(b,x^4\cdot3^5=27^3\)
\(\Rightarrow x^4\cdot3^5=\left(3^3\right)^3\)
\(\Rightarrow x^4\cdot3^5=3^9\)
\(\Rightarrow x^4=3^9:3^5\)
\(\Rightarrow x^4=3^4\)
\(\Rightarrow x=3\)
Vậy \(x=3\)
\(c,\dfrac{8}{3}\cdot\left(\dfrac{5}{24}-x\right)=\dfrac{-1}{3}\)
\(\Rightarrow\dfrac{5}{24}-x=\dfrac{-1}{3}:\dfrac{8}{3}\)
\(\Rightarrow\dfrac{5}{24}-x=\dfrac{-1}{8}\)
\(\Rightarrow x=\dfrac{5}{24}-\dfrac{-1}{8}\)
\(\Rightarrow x=\dfrac{5}{24}+\dfrac{1}{8}\)
\(\Rightarrow x=\dfrac{1}{3}\)
Vậy \(x=\dfrac{1}{3}\)