Để \(\dfrac{x^3-16x}{x^3-3x^2-4x}=1\) thì
x3-16x=x3-3x2-4x
=>x3-16x-x3+3x2+4x=0
=>3x2-12x=0
=>3x(x-4)=0
=>x-4=0
=>x=4
vậy x=4
\(\dfrac{x^3-16}{x^3-3x^2-4x}=1\)
\(\Leftrightarrow\dfrac{x^3-16}{x^3-3x^2-4x}-1=0\)
\(\Leftrightarrow\dfrac{x^3-16}{x^3-3x^2-4x}-\dfrac{x^3-3x^2-4x}{x^3-3x^2-4x}=0\)
\(\Leftrightarrow\dfrac{x^3-16-\left(x^3-3x^2-4x\right)}{x^3-3x^2-4x}=0\)
\(\Leftrightarrow x^3-16-\left(x^3-3x^2-4x\right)=0\)
\(\Leftrightarrow x^3-16-x^3+3x^2+4x=0\)
\(\Leftrightarrow3x^2+4x-16=0\)
\(\Leftrightarrow x=1,737034184\)
\(\dfrac{x^3-16x}{x^3-3x^2-4x}=1\)
\(\Leftrightarrow\dfrac{x^3-16x}{x^3-3x^2-4x}-1=0\)
\(\Leftrightarrow\dfrac{x^3-16x}{x^3-3x^2-4x}-\dfrac{x^3-3x^2-4x}{x^3-3x^2-4x}=0\)
\(\Leftrightarrow\dfrac{x^3-16x-\left(x^3-3x^2-4x\right)}{x^3-3x^2-4x}=0\)
\(\Leftrightarrow x^3-16x-\left(x^3-3x^2-4x\right)=0\)
\(\Leftrightarrow x^3-16x-x^3+3x^2+4x=0\)
\(\Leftrightarrow3x^2-12x=0\)
\(\Leftrightarrow3x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy x=0 hoặc x=4