Tìm x, biết: a) 50%x + 0,75x = x - 5 b) ( $x^{2}$ - 4 )( 2x + $\frac{3}{5}$ ) = 0 c) -$\frac{4}{7}$$x^{2}$ + 3x = 0

#andy

\[\begin{array}{l}
a)50\% x + 0,75x = x - 5\\
 \Leftrightarrow \dfrac{1}{2}x + 0,75x = x - 5\\
 \Leftrightarrow 1,25x = x - 5\\
 \Leftrightarrow 1,25x - 1x =  - 5\\
 \Leftrightarrow 0,25x =  - 5\\
 \Leftrightarrow x =  - 20\\
b)\left( {{x^2} - 4} \right)\left( {2x + \dfrac{3}{5}} \right) = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left( {x + 2} \right)\left( {2x + \dfrac{3}{5}} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
x + 2 = 0\\
2x + \dfrac{3}{5} = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x =  - 2\\
2x =  - \dfrac{3}{5}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x =  - 2\\
x =  - \dfrac{3}{{10}}
\end{array} \right.\\
 \Rightarrow S = \left\{ {2; - 2; - \dfrac{3}{{10}}} \right\}\\
c) - \dfrac{4}{7}{x^2} + 3x = 0\\
 \Leftrightarrow x.\left( { - \dfrac{4}{7}x + 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
 - \dfrac{4}{7}x + 3 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
 - \dfrac{4}{7}x =  - 3
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \dfrac{{21}}{4}
\end{array} \right.\\
 \Rightarrow S = \left\{ {0;\dfrac{{21}}{4}} \right\}
\end{array}\]