\(ĐKXĐ:x\ne0;x\ne2\)
\(\frac{4x^2-4x^3+x^4}{x^3-2x^2}=-2\)
\(\Leftrightarrow4x^2-4x^3+x^4=-2\left(x^3-2x^2\right)\)
\(\Leftrightarrow4x^2-4x^3+x^4=-2x^3+4x^2\)
\(\Leftrightarrow x^4-2x^3=0\Leftrightarrow x^3\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\left(ktm\right)\)
Vậy không có x để phân thức bằng -2
Ta có : \(\frac{4x^2-4x^3+x^4}{x^3-2x^2}=-2\)
( ĐKXĐ : \(x\ne0,x\ne\pm\sqrt{2}\) )
\(\Leftrightarrow\frac{4x^2-4x^3+x^4}{x^3-2x^2}+2=0\)
\(\Leftrightarrow4x^2-4x^3+x^4+2\left(x^3-2x^2\right)=0\)
\(\Leftrightarrow-2x^3+x^4=0\)
\(\Leftrightarrow x^3\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\) ( Loại \(x=0\) không thỏa mãn ĐKXĐ )
Vậy : \(x=2\) thỏa mãn đề.
ĐKXĐ: \(x\ne0;2\)
\(\frac{4x^2-4x^3+x^4}{x^3-2x^2}=-2\)
\(\Leftrightarrow\frac{x^2\left(4-4x+x^2\right)}{x^3-2x^2}=-2\)
\(\Leftrightarrow\frac{x^2\left(2-x^2\right)^2}{x^2\left(x-2\right)}=-2\)
\(\Leftrightarrow\frac{\left(2-x\right)^2}{x-2}=-2\)
\(\Leftrightarrow\left(2-x\right)^2=-2\left(x-2\right)\)
\(\Leftrightarrow4-4x+x^2=-2x+4\)
\(\Leftrightarrow-4x+x^2=-2x\)
\(\Leftrightarrow4x-x^2-2x=0\)
\(\Leftrightarrow2x-x^2=0\)
\(\Leftrightarrow x\left(2-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=2\left(ktm\right)\end{cases}}\)
