a) Ta có: \(\dfrac{1}{5}x+2=0\)
nên \(\dfrac{1}{5}x=-2\)
hay x=-10
b) Ta có: \(\left(2x+4\right)\left(6-\dfrac{2}{3}x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+4=0\\6-\dfrac{2}{3}x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-4\\\dfrac{2}{3}x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=9\end{matrix}\right.\)
c) Ta có: \(\dfrac{1}{4}x^3+9x=0\)
\(\Leftrightarrow x\left(\dfrac{1}{4}x^2+9\right)=0\)
mà \(\dfrac{1}{4}x^2+9>0\forall x\)
nên x=0
d) Ta có: \(x^2+4x-5=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
