\(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)+\dfrac{5}{9}=\dfrac{23}{27}\)
=> \(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)=\dfrac{23}{27}-\dfrac{5}{9}\)
=> \(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)=\dfrac{23}{27}-\dfrac{15}{27}\)
=> \(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)=\dfrac{8}{27}\)
=> \(2+\dfrac{3}{4}x=\dfrac{7}{9}:\dfrac{8}{27}\)
=> \(2+\dfrac{3}{4}x=\dfrac{7}{9}.\dfrac{27}{8}\)
=> \(2+\dfrac{3}{4}x=\dfrac{7.3}{1.8}\)
=> \(2+\dfrac{3}{4}x=\dfrac{21}{8}\)
=> \(\dfrac{3}{4}x=\dfrac{21}{8}-2\)
=> \(\dfrac{3}{4}x=\dfrac{21}{8}-\dfrac{16}{8}\)
=> \(\dfrac{3}{4}x=\dfrac{5}{8}\)
=> \(x=\dfrac{5}{8}:\dfrac{3}{4}\)
=> \(x=\dfrac{5}{8}.\dfrac{4}{3}\)
=> \(x=\dfrac{5}{6}\)
Vậy x = \(\dfrac{5}{6}\)
