Question

\(\dfrac{1}{2}x+\dfrac{3}{5}x=\dfrac{-22}{25}\)

\(\dfrac{6}{7}x-\dfrac{1}{2}x=\dfrac{15}{28}\)

\(4x\cdot\left(x+\dfrac{3}{5}\right)=0\) 

\(\left(\dfrac{5}{8}+x\right)\left(3x-\dfrac{6}{11}\right)=0\)

 

Answer 1

1, \(\dfrac{x}{2}+\dfrac{3x}{5}=-\dfrac{22}{25}\Leftrightarrow25x+30x=-44\Leftrightarrow55x=-44\Leftrightarrow x=-\dfrac{44}{55}=-\dfrac{4}{5}\)

2, \(\dfrac{6x}{7}-\dfrac{x}{2}=\dfrac{15}{28}\Leftrightarrow24x-14x=15\Leftrightarrow10x=15\Leftrightarrow x=\dfrac{3}{2}\)

3, \(4x\left(x+\dfrac{3}{5}\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{5}\end{matrix}\right.\)

4, \(\left(\dfrac{5}{8}+x\right)\left(3x-\dfrac{6}{11}\right)=0\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{5}{8}=0\\3x-\dfrac{6}{11}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{8}\\x=\dfrac{6}{11}:3=\dfrac{6}{33}=\dfrac{2}{11}\end{matrix}\right.\)