a) \(x+\dfrac{5}{6}=\dfrac{4}{3}\)                  

\(x=\dfrac{4}{3}-\dfrac{5}{6}\)

\(x=\dfrac{8}{6}-\dfrac{5}{6}\)

\(x=\dfrac{1}{2}\)

Vậy x = \(\dfrac{1}{2}\)

b) \(x:2^4=8^3\)

\(x=8^3.2^4\)

\(x=\left(2^3\right)^3.2^4\)

\(x=2^9.2^4\)

\(x=\left(2\right)^{9+4}\)

\(x=2^{13}\)

Vậy x = \(2^{13}\)

c) \(\dfrac{13}{4}.\left(\dfrac{5}{52}-x\right)=\dfrac{1}{4}\)

\(\left(\dfrac{5}{52}-x\right)=\dfrac{1}{4}:\dfrac{13}{4}\)

\(\dfrac{5}{52}-x=\dfrac{1}{4}.\dfrac{4}{13}\)

\(\dfrac{5}{52}-x=\dfrac{1}{13}\)

\(x=\dfrac{5}{52}-\dfrac{1}{13}\)

\(x=\dfrac{5}{52}-\dfrac{4}{52}\)

\(x=\dfrac{1}{52}\)

Vậy x = \(\dfrac{1}{52}\)

a) \(x+\dfrac{5}{6}=\dfrac{4}{3}\)                  

\(x=\dfrac{4}{3}-\dfrac{5}{6}\)

\(x=\dfrac{8}{6}-\dfrac{5}{6}\)

\(x=\dfrac{1}{2}\)

Vậy x = \(\dfrac{1}{2}\)

b) \(x:2^4=8^3\)

\(x=8^3.2^4\)

\(x=\left(2^3\right)^3.2^4\)

\(x=2^9.2^4\)

\(x=\left(2\right)^{9+4}\)

\(x=2^{13}\)

Vậy x = \(2^{13}\)

c) \(\dfrac{13}{4}.\left(\dfrac{5}{52}-x\right)=\dfrac{1}{4}\)

\(\left(\dfrac{5}{52}-x\right)=\dfrac{1}{4}:\dfrac{13}{4}\)

\(\dfrac{5}{52}-x=\dfrac{1}{4}.\dfrac{4}{13}\)

\(\dfrac{5}{52}-x=\dfrac{1}{13}\)

\(x=\dfrac{5}{52}-\dfrac{1}{13}\)

\(x=\dfrac{5}{52}-\dfrac{4}{52}\)

\(x=\dfrac{1}{52}\)

Vậy x = \(\dfrac{1}{52}\)

a) \(x+\dfrac{5}{6}=\dfrac{4}{3}\)                  

\(x=\dfrac{4}{3}-\dfrac{5}{6}\)

\(x=\dfrac{8}{6}-\dfrac{5}{6}\)

\(x=\dfrac{1}{2}\)

Vậy x = \(\dfrac{1}{2}\)

b) \(x:2^4=8^3\)

\(x=8^3.2^4\)

\(x=\left(2^3\right)^3.2^4\)

\(x=2^9.2^4\)

\(x=\left(2\right)^{9+4}\)

\(x=2^{13}\)

Vậy x = \(2^{13}\)

c) \(\dfrac{13}{4}.\left(\dfrac{5}{52}-x\right)=\dfrac{1}{4}\)

\(\left(\dfrac{5}{52}-x\right)=\dfrac{1}{4}:\dfrac{13}{4}\)

\(\dfrac{5}{52}-x=\dfrac{1}{4}.\dfrac{4}{13}\)

\(\dfrac{5}{52}-x=\dfrac{1}{13}\)

\(x=\dfrac{5}{52}-\dfrac{1}{13}\)

\(x=\dfrac{5}{52}-\dfrac{4}{52}\)

\(x=\dfrac{1}{52}\)

Vậy x = \(\dfrac{1}{52}\)

a) \(\dfrac{5}{9}-\left(\dfrac{1}{3}^{ }\right)^2_{ }=\dfrac{5}{9}-\dfrac{1}{9}=\dfrac{4}{9}\)

b) \(\dfrac{1}{5}.\dfrac{-3}{2}+\dfrac{-17}{2}.\dfrac{1}{5}=\dfrac{1}{5}.\left(\dfrac{-3}{2}+\dfrac{-17}{2}\right)=\dfrac{1}{5}.-10=-2\)

c)\(1+\left(\dfrac{-2}{5}+\dfrac{11}{13}\right)-\left(\dfrac{3}{5}-\dfrac{2}{13}\right)=1+\dfrac{-2}{5}+\dfrac{11}{13}-\dfrac{3}{5}+\dfrac{2}{13}=1+\left(\dfrac{-2}{5}-\dfrac{3}{5}\right)+\left(\dfrac{11}{13}+\dfrac{2}{13}\right)_{ }=1+-1+1=1\)