$ĐKXĐ : x \neq 2, x \neq 0, x \neq -2$
Ta có : $\dfrac{4x-2x^2}{x^3-4x} = \dfrac{1}{2}$
$\to \dfrac{2x.(x-2)}{x.(x-2).(x+2)} = \dfrac{1}{2}$
$\to \dfrac{2}{x+2} = \dfrac{1}{2} = \dfrac{2}{4}$
$\to x+2=4$
$\to x=2$ ( Không thỏa mãn $ĐKXĐ$ )
Vậy pt vô nghiệm.
\(\frac{4x-2x^2}{x^3-4x}=\frac{1}{2}\)
ĐKXĐ: \(x^3-4x\ne0\Leftrightarrow x\notin\left\{0;2;-2\right\}\)
\(pt\Leftrightarrow2\left(4x-2x^2\right)=x^3-4x\\ \Leftrightarrow8x-4x^2-x^3+4x=0\\ \Leftrightarrow-x^3-4x^2+12x=0\\ \Leftrightarrow-x^3-6x^2+2x^2+12x=0\\ \Leftrightarrow-x^2\left(x+6\right)+2x\left(x+6\right)=0\\ \Leftrightarrow\left(-x^2+2x\right)\left(x+6\right)=0\\ \Leftrightarrow-x\left(x-2\right)\left(x+6\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(ktm\right)\\x=-6\left(tmđk\right)\end{matrix}\right.\)
Vậy pt ban đầu có tập nghiệm \(S=\left\{-6\right\}\)
