Bài 1:
a) ĐKXĐ: \(9x^2-6x+1\neq 0\)
\(\Leftrightarrow (3x-1)^2\neq 0\Leftrightarrow x\neq \frac{1}{3}\)
b) Với \(x=-8\Rightarrow C=\frac{3(-8)^2-(-8)}{9(-8)^2-6(-8)+1}=\frac{8}{25}\)
c) Ta có:
\(C=\frac{3x^2-x}{9x^2-6x+1}=\frac{x(3x-1)}{(3x-1)^2}=\frac{x}{3x-1}\)
d)
Phân thức đã cho nhận giá trị âm \(\Leftrightarrow \) \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\3x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\3x-1< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow \) \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x>\dfrac{1}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x< \dfrac{1}{3}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x>0\\ x< \frac{1}{3}\end{matrix}\right.\)
Bài 2:
a) ĐKXĐ: \((x+1)(2x-6)\neq 0\)
\(\Leftrightarrow \left\{\begin{matrix} x+1\neq 0\\ 2x-6\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq -1\\ x\neq 3\end{matrix}\right.\)
b) Ta có:
\(\frac{3x^2+3x}{(x+1)(2x-6)}=1\)
\(\Leftrightarrow \frac{3x(x+1)}{(x+1)(2x-6)}=1\)
\(\Leftrightarrow \frac{3x}{2x-6}=1\Leftrightarrow 3x=2x-6\Leftrightarrow x=-6\)
c) Để phân thức đã cho nhận giá trị dương thì:
\(\frac{3x}{2x-6}>0\Leftrightarrow \frac{x}{x-3}>0\)
\(\Leftrightarrow \)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x-3< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow \) \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x>3\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x< 3\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x>3\\x< 0\end{matrix}\right.\)
Vậy để pt nhận giá trị dương thì \(x\neq -1; x\neq 3\) và \(x>3\) hoặc \(x<0\)