2:
a: \(\dfrac{2}{5}x-\dfrac{1}{3}=\dfrac{4}{9}\)
=>\(\dfrac{2}{5}x=\dfrac{4}{9}+\dfrac{1}{3}=\dfrac{4}{9}+\dfrac{3}{9}=\dfrac{7}{9}\)
=>\(x=\dfrac{7}{9}:\dfrac{2}{5}=\dfrac{7}{9}\cdot\dfrac{5}{2}=\dfrac{35}{18}\)
b: \(\left(x-\dfrac{1}{2}\right)\left(5x-\dfrac{1}{3}\right)=0\)
=>\(\left[{}\begin{matrix}x-\dfrac{1}{2}=0\\5x-\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{15}\end{matrix}\right.\)
c: \(\left|x-\dfrac{2}{5}\right|=\dfrac{1}{4}\)
=>\(\left[{}\begin{matrix}x-\dfrac{2}{5}=\dfrac{1}{4}\\x-\dfrac{2}{5}=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}+\dfrac{2}{5}=\dfrac{5+8}{20}=\dfrac{13}{20}\\x=\dfrac{-1}{4}+\dfrac{2}{5}=\dfrac{-5+8}{20}=\dfrac{3}{20}\end{matrix}\right.\)
1:
a: \(\left(\dfrac{1}{3}\right)^2\cdot\sqrt{9}+\left|-\dfrac{2}{3}\right|=\dfrac{1}{9}\cdot3+\dfrac{2}{3}\)
\(=\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{3}{3}=1\)
b: \(25,3+\left(-4,8\right)-5,3+4,8\)
=25,3-5,3+4,8-4,8
=20+0
=20
c: \(\dfrac{2}{7}\cdot\dfrac{-6}{11}+\dfrac{2}{7}\cdot\dfrac{-5}{11}\)
\(=\dfrac{2}{7}\left(-\dfrac{6}{11}-\dfrac{5}{11}\right)=-\dfrac{2}{7}\)
