Câu 3:
a: \(4,5-\left(2x-\dfrac{1}{3}\right)=3,5\)
=>\(2x-\dfrac{1}{3}=4,5-3,5=1\)
=>\(2x=1+\dfrac{1}{3}=\dfrac{4}{3}\)
=>\(x=\dfrac{2}{3}\)
b: \(\left|2\dfrac{1}{2}-x\right|:3=0,5\)
=>\(\left|x-\dfrac{5}{2}\right|=0,5\cdot3=1,5=\dfrac{3}{2}\)
=>\(\left[{}\begin{matrix}x-\dfrac{5}{2}=\dfrac{3}{2}\\x-\dfrac{5}{2}=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}+\dfrac{5}{2}=\dfrac{8}{2}=4\\x=-\dfrac{3}{2}+\dfrac{5}{2}=\dfrac{2}{2}=1\end{matrix}\right.\)
c: \(\dfrac{3}{2}-2x=\left|-\dfrac{1}{4}\right|\)
=>\(\dfrac{3}{2}-2x=\dfrac{1}{4}\)
=>\(2x=\dfrac{3}{2}-\dfrac{1}{4}=\dfrac{5}{4}\)
=>\(x=\dfrac{5}{8}\)
d: \(\left(x+\dfrac{3}{5}\right)-\dfrac{1}{2}=\sqrt{\dfrac{16}{9}}\)
=>\(x+\dfrac{3}{5}-\dfrac{1}{2}=\dfrac{4}{3}\)
=>\(x+\dfrac{1}{10}=\dfrac{4}{3}\)
=>\(x=\dfrac{4}{3}-\dfrac{1}{10}=\dfrac{37}{30}\)