Câu 2:
a)$\frac{2}{3}$× x - $\frac{4}{5}$ =$\frac{1}{5}$
= $\frac{2}{3}$ × x =$\frac{1}{5}$ +$\frac{4}{5}$
=$\frac2{}{3}$ × x =1
x =1÷$\frac{2}{3}$
x =$\frac{3}{2}$
b) |x+$\frac{2}{3}$|-$\frac{6}{9}$ =$\frac{7}{3}$
|x+$\frac{2}{3}$| =$\frac{7}{3}$+$\frac{6}{9}$
|x+$\frac{2}{3}$| =3
x+$\frac{2}{3}$ =3 hoặc x+$\frac{2}{3}$ =-3
x =3-$\frac{2}{3}$ hoặc x =-3-$\frac{2}{3}$
x =$\frac{7}{3}$ hoặc x =$\frac{11}{3}$
`c) 8^x = 64`
`=> 8^x = 8^2`
`=> x = 2`
