a) Ta có: \(\frac{4}{5}+x=\frac{2}{3}\)
hay \(x=\frac{2}{3}-\frac{4}{5}=\frac{10}{15}-\frac{12}{15}=\frac{-2}{15}\)
Vậy: \(x=\frac{-2}{15}\)
b) Ta có: \(\frac{3}{4}-x=\frac{1}{3}\)
hay \(x=\frac{3}{4}-\frac{1}{3}=\frac{9}{12}-\frac{4}{12}=\frac{5}{12}\)
Vậy: \(x=\frac{5}{12}\)
c) Ta có: \(x-\frac{5}{9}=\frac{-2}{3}\)
hay \(x=\frac{-2}{3}+\frac{5}{9}=\frac{-6}{9}+\frac{5}{9}=-\frac{1}{9}\)
Vậy: \(x=\frac{-1}{9}\)
d) Ta có: \(\frac{\left(4,5-2x\right)}{\frac{3}{4}}=1\frac{1}{3}\)
\(\Leftrightarrow\frac{9}{2}-2x=\frac{4}{3}\cdot\frac{3}{4}=1\)
\(\Leftrightarrow2x=\frac{9}{2}-1=\frac{7}{2}\)
hay \(x=\frac{7}{4}\)
Vậy: \(x=\frac{7}{4}\)
