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1.
\(f\left(x\right)=\frac{x-7}{\left(x-4\right)\left(4x-3\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\left\{\frac{3}{4};4\right\}\)
\(f\left(x\right)=0\Rightarrow x=7\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}\frac{3}{4}< x< 4\\x>7\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{3}{4}\\4< x< 7\end{matrix}\right.\)
2.
\(f\left(x\right)=\frac{11x+3}{-\left(x-\frac{5}{2}\right)^2-\frac{3}{4}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=-\frac{3}{11}\)
\(f\left(x\right)>0\Rightarrow x< -\frac{3}{11}\)
\(f\left(x\right)< 0\Rightarrow x>-\frac{3}{11}\)
3.
\(f\left(x\right)=\frac{3x-2}{\left(x-1\right)\left(x^2-2x-2\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định khi \(x=\left\{1;1\pm\sqrt{3}\right\}\)
\(f\left(x\right)=0\Rightarrow x=\frac{2}{3}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< 1-\sqrt{3}\\\frac{2}{3}< x< 1\\x>1+\sqrt{3}\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}1-\sqrt{3}< x< \frac{2}{3}\\1< x< 1+\sqrt{3}\end{matrix}\right.\)
4.
\(f\left(x\right)=\frac{\left(x-2\right)\left(x+6\right)}{\sqrt{6}\left(x+\frac{\sqrt{6}}{4}\right)^2+\frac{8\sqrt{2}-3\sqrt{6}}{8}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=\left\{-6;2\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -6\\x>2\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow-6< x< 2\)
a/ \(\left(2x-3\right)\left(3x-4\right)\left(5x+2\right)>0\)
\(\Rightarrow\left[{}\begin{matrix}-\frac{2}{3}< x< \frac{4}{3}\\x>\frac{3}{2}\end{matrix}\right.\)
b/ \(\Leftrightarrow24x^2-10x-25< 0\)
\(\Rightarrow-\frac{5}{6}< x< \frac{5}{4}\)
c/ \(\frac{4x\left(3x+2\right)}{2x+5}>0\Rightarrow\left[{}\begin{matrix}-\frac{5}{2}< x< -\frac{2}{3}\\x>0\end{matrix}\right.\)
d/ \(\Leftrightarrow\frac{3x+2}{2x-5}-\frac{2x-5}{3x+2}\ge0\)
\(\Leftrightarrow\frac{\left(3x+2\right)^2-\left(2x-5\right)^2}{\left(2x-5\right)\left(3x+2\right)}\ge0\)
\(\Leftrightarrow\frac{\left(5x-2\right)\left(x+7\right)}{\left(2x-5\right)\left(3x+2\right)}\ge0\Rightarrow\left[{}\begin{matrix}x\le-7\\-\frac{2}{3}< x\le\frac{2}{5}\\x>\frac{5}{2}\end{matrix}\right.\)