Cho

D

- Hướng dẫn giải

Đáp án D

$$\begin{gathered} +) \frac{1}{2}-\left(\frac{2}{3}x-\frac{1}{3}\right)=\frac{-2}{3} \\ \frac{2}{3}x-\frac{1}{3}=\frac{1}{2}-\left(\frac{-2}{3}\right) \\ \frac{2}{3}x-\frac{1}{3}=\frac{7}{6} \\ \frac{2}{3}x=\frac{7}{6}+\frac{1}{3} \\ \frac{2}{3}x=\frac{3}{2} \\ x=\frac{3}{2}:\frac{2}{3} \\ x=\frac{9}{4} \end{gathered}$$

Nên $x_1=\frac{9}{4}$

$$\begin{gathered} +) \frac{5}{6}-x=\frac{-1}{12}+\frac{4}{3} \\ \frac{5}{6}-x=\frac{5}{4} \\ x=\frac{5}{6}-\frac{5}{4} \\ x=\frac{-5}{12} \end{gathered}$$

Nên $x_2=-\frac{5}{12}$

Từ đó $x_1+x_2=\frac{9}{4}+\left(\frac{-5}{12}\right)=\frac{11}{6}$