a) \(x^3+4x=0\)
\(\Rightarrow x\left(x^2+4\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x^2+4=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x^2=-4\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x\in\phi\end{array}\right.\)
Vậy: \(x=0\)
b) \(2\left(5-x\right)=4x-3\)
\(\Rightarrow10-2x=4x-3\)
\(\Rightarrow10+3=4x+2x\)
\(\Rightarrow13=6x\)
\(\Rightarrow x=\frac{13}{6}\)
x3+ 4x=0
<=> x(x2+4)=0
=> x=0 hoặc x2+4=0
Mà: x2+4 >4
=>x=0
b) (5-x).2=4x-3
<=>10-2x=4x-3
<=>10+3= 2x+4x
<=>13=6x
=>x=\(\frac{13}{6}\)