Bài 1: 

a) \(1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{6}{6}-\dfrac{3}{6}+\dfrac{2}{6}=\dfrac{5}{6}\)

b) \(\dfrac{2}{5}+\dfrac{3}{5}:\dfrac{9}{10}=\dfrac{2}{5}+\dfrac{3}{5}\cdot\dfrac{10}{9}=\dfrac{2}{5}+\dfrac{2}{3}=\dfrac{16}{15}\)

Bài 1:

c) \(\dfrac{7}{11}\cdot\dfrac{3}{4}+\dfrac{7}{11}\cdot\dfrac{1}{4}+\dfrac{4}{11}=\dfrac{7}{11}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\dfrac{4}{11}=\dfrac{7}{11}+\dfrac{4}{11}=1\)

d) \(\left(\dfrac{3}{4}+0.5+25\%\right)\cdot2\dfrac{2}{3}=\dfrac{3}{2}\cdot\dfrac{8}{3}=4\)

Bài 2:

a) Ta có: \(x+\dfrac{1}{2}=2\)

\(\Leftrightarrow x=2-\dfrac{1}{2}=\dfrac{3}{2}\)

Vậy: \(x=\dfrac{3}{2}\)

Bài 2: 

b) Ta có: \(75\%\cdot x-\dfrac{1}{2}=-1\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{3}{4}x=\dfrac{-5}{4}+\dfrac{1}{2}=\dfrac{-5}{4}+\dfrac{2}{4}=\dfrac{-3}{4}\)

hay x=-1

Vậy: x=-1

Bài 2: 

c) Ta có: \(\left|x-\dfrac{2}{5}\right|+\dfrac{7}{10}=3\)

\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=3-\dfrac{7}{10}=\dfrac{23}{10}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=\dfrac{23}{10}\\x-\dfrac{2}{5}=\dfrac{-23}{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{23}{10}+\dfrac{2}{5}=\dfrac{27}{10}\\x=\dfrac{-23}{10}+\dfrac{2}{5}=\dfrac{-19}{10}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{27}{10};-\dfrac{19}{10}\right\}\)