Bài 3 : Xét dấu biểu thức sau :
1 , \(f\left(x\right)=\frac{x-7}{4x^2-19x+12}\)
2 , \(f\left(x\right)=\frac{11x+3}{-x^2+5x-7}\)
3 , \(f\left(x\right)=\frac{3x-2}{x^3-3x^2+2}\)
4 , \(f\left(x\right)=\frac{x^2+4x-12}{\sqrt{6}x^2+3x+\sqrt{2}}\)
5 , \(f\left(x\right)=\frac{x^2-3x-2}{-x^2+x-1}\)
6 , \(f\left(x\right)=\frac{x^3-5x+4}{x^4-4x^3+8x-5}\)
7 , \(f\left(x\right)=\frac{\left(x+3\right)\left(x-2\right)\left(-2x^2+x-1\right)}{\left(2x-5\right)\left(x^2+3x-10\right)}\)
8 , \(f\left(x\right)=\left(-x^2+x-1\right)\left(6x^2-5x+1\right)\)
9 , \(f\left(x\right)=\frac{x^2-x-2}{-x^2+3x+4}\)
10 , \(f\left(x\right)=\left(x^2-5x+4\right)\left(2-5x+2x^2\right)\)
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:
\(f\left(x\right)=0\Rightarrow x=\left\{-6;2\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -6\\x>2\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow-6< x< 2\)
5.
\(f\left(x\right)=\frac{x^2-3x-2}{-\left(x-\frac{1}{2}\right)^2-\frac{3}{4}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=\frac{3\pm\sqrt{17}}{2}\)
\(f\left(x\right)>0\Rightarrow\frac{3-\sqrt{17}}{2}< x< \frac{3+\sqrt{17}}{2}\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{3-\sqrt{17}}{2}\\x>\frac{3+\sqrt{17}}{2}\end{matrix}\right.\)
6.
\(f\left(x\right)=\frac{\left(x-1\right)\left(x^2+x-4\right)}{\left(x-1\right)^2\left(x^2-2x-5\right)}=\frac{x^2+x-4}{\left(x-1\right)\left(x^2-2x-5\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định khi \(x=\left\{1;1\pm\sqrt{6}\right\}\)
\(f\left(x\right)=0\Rightarrow x=\left\{\frac{-1\pm\sqrt{17}}{2}\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}\frac{-1-\sqrt{17}}{2}< x< 1-\sqrt{6}\\1< x< \frac{-1+\sqrt{17}}{2}\\x>1+\sqrt{6}\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{-1-\sqrt{17}}{2}\\1-\sqrt{6}< x< 1\\\frac{-1+\sqrt{17}}{2}< x< 1+\sqrt{6}\end{matrix}\right.\)
7.
\(f\left(x\right)=\frac{\left(x+3\right)\left(x-2\right)\left[-2\left(x-\frac{1}{4}\right)^2-\frac{7}{8}\right]}{\left(2x-5\right)\left(x-2\right)\left(x+5\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\left\{-5;2;\frac{5}{2}\right\}\)
\(f\left(x\right)=0\Rightarrow x=-3\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -5\\-3< x< 2\\2< x< \frac{5}{2}\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}-5< x< -3\\x>\frac{5}{2}\end{matrix}\right.\)
8.
\(f\left(x\right)=-\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]\left(2x-1\right)\left(3x-1\right)\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=\left\{\frac{1}{3};\frac{1}{2}\right\}\)
\(f\left(x\right)>0\Rightarrow\frac{1}{3}< x< \frac{1}{2}\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x>\frac{1}{2}\\x< \frac{1}{3}\end{matrix}\right.\)
9.
\(f\left(x\right)=\frac{\left(x+1\right)\left(x-2\right)}{\left(x+1\right)\left(4-x\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\left\{-1;4\right\}\)
\(f\left(x\right)=0\Rightarrow x=2\)
\(f\left(x\right)>0\Rightarrow2< x< 4\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< -1\\-1< x< 2\\x>4\end{matrix}\right.\)
10.
\(f\left(x\right)=\left(x-1\right)\left(x-4\right)\left(2x-1\right)\left(x-2\right)\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=\left\{\frac{1}{2};1;2;4\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< \frac{1}{2}\\1< x< 2\\x>4\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}\frac{1}{2}< x< 1\\2< x< 4\end{matrix}\right.\)