câu 1.
\(\dfrac{2}{3}-\dfrac{3}{6}=\dfrac{4}{6}-\dfrac{3}{6}=\dfrac{1}{6}\)
\(\dfrac{3}{5}+-\dfrac{15}{7}+\dfrac{7}{5}+-\dfrac{6}{7}=\dfrac{3}{5}+\dfrac{7}{5}+\left(\dfrac{-15}{7}-\dfrac{6}{7}\right)=2+-3=-1\)
Câu 2:
a) Ta có: \(\dfrac{1}{2}x-\dfrac{1}{4}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{1}{2}x=\dfrac{1}{6}+\dfrac{1}{4}=\dfrac{5}{12}\)
\(\Leftrightarrow x=\dfrac{5}{12}:\dfrac{1}{2}=\dfrac{5}{12}\cdot\dfrac{2}{1}=\dfrac{5}{6}\)
Vậy: \(x=\dfrac{5}{6}\)
b) Ta có: \(\dfrac{1}{4}+75\%\cdot y=\dfrac{-1}{2}\)
\(\Leftrightarrow\dfrac{3}{4}\cdot y=\dfrac{-1}{2}-\dfrac{1}{4}=\dfrac{-2}{4}-\dfrac{1}{4}=\dfrac{-3}{4}\)
hay y=-1
Vậy: y=-1
2.
a) \(\dfrac{1}{2}x-\dfrac{1}{4}=\dfrac{1}{6}\Leftrightarrow\dfrac{1}{2}x=\dfrac{1}{6}-\dfrac{1}{4}\Leftrightarrow\dfrac{1}{2}x=\dfrac{1}{-12}:\dfrac{1}{2}=-\dfrac{1}{6}\)
