Đặt \(A=\dfrac{3x^2-x}{9x^2-6x+1}\)
a)
A xác định \(\Leftrightarrow9x^2-6x+1\ne0\Leftrightarrow\left(3x-1\right)^2\ne0\Leftrightarrow x\ne\dfrac{1}{3}\)
b)
x=-8
\(A=\dfrac{3x^2-x}{9x^2-6x+1}=\dfrac{x\left(3x-1\right)}{\left(3x-1\right)^2}=\dfrac{x}{3x-1}=-\dfrac{8}{3.\left(-8\right)-1}=\dfrac{8}{25}\)
c)
\(A=\dfrac{3x^2-x}{9x^2-6x+1}=\dfrac{x\left(3x-1\right)}{\left(3x-1\right)^2}=\dfrac{x}{3x-1}\)
d)
A âm \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\3x-1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\3x-1>0\end{matrix}\right.\end{matrix}\right.\)
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