Bài 4 Xét dấu biểu thức sau
1 , \(f\left(x\right)=x^2-3x-2-\frac{8}{x^2-3x}\)
2 , \(f\left(x\right)=\frac{1}{x+1}-\frac{1}{x}-\frac{1}{2}\)
3 , \(f\left(x\right)=\frac{x^2-4x+3}{3-2x}-1+x\)
4 , \(f\left(x\right)=\frac{x^2-1}{\left(x^2-3\right)\left(-3x^2+2x+8\right)}\)
5 , \(f\left(x\right)=x^4-5x^2+2x+3\)
6 , \(f\left(x\right)=\frac{x^2+4x+15}{x^2-1}-\frac{x-3}{x+1}-\frac{x-2}{1-x}\)
1.
\(f\left(x\right)=\frac{\left(x^2-3x\right)^2-2\left(x^2-3x\right)-8}{x^2-3x}=\frac{\left(x^2-3x-4\right)\left(x^2-3x+2\right)}{x^2-3x}\)
\(f\left(x\right)=\frac{\left(x+1\right)\left(x-1\right)\left(x-2\right)\left(x-4\right)}{x\left(x-3\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\left\{0;3\right\}\)
\(f\left(x\right)=0\Rightarrow x=\left\{-1;1;2;4\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -1\\0< x< 1\\2< x< 3\\x>4\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}-1< x< 0\\1< x< 2\\3< x< 4\end{matrix}\right.\)
2.
\(f\left(x\right)=\frac{2x-2\left(x+1\right)-x\left(x+1\right)}{2x\left(x+1\right)}=\frac{-x^2-x-2}{2x\left(x+1\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\left\{-1;0\right\}\)
\(f\left(x\right)>0\Rightarrow-1< x< 0\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< -1\\x>0\end{matrix}\right.\)
3.
\(f\left(x\right)=\frac{x^2-4x+3+\left(x-1\right)\left(3-2x\right)}{3-2x}=\frac{-x^2+x}{3-2x}=\frac{x\left(1-x\right)}{3-2x}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\frac{3}{2}\)
\(f\left(x\right)=0\Rightarrow x=\left\{0;1\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}0< x< 1\\x>\frac{3}{2}\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< 0\\1< x< \frac{3}{2}\end{matrix}\right.\)
4.
\(f\left(x\right)=\frac{\left(x-1\right)\left(x+1\right)}{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\left(2-x\right)\left(3x+4\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\left\{\pm\sqrt{3};-\frac{4}{3};2\right\}\)
\(f\left(x\right)=0\Rightarrow x=\pm1\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}-\sqrt{3}< x< -\frac{4}{3}\\-1< x< 1\\\sqrt{3}< x< 2\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< -\sqrt{3}\\-\frac{4}{3}< x< -1\\1< x< \sqrt{3}\\x>2\end{matrix}\right.\)
5.
\(f\left(x\right)=x^4-x^3-x^2+x^3-x^2-x-3x^2+3x+3\)
\(=x^2\left(x^2-x-1\right)+x\left(x^2-x-1\right)-3\left(x^2-x-1\right)\)
\(=\left(x^2+x-3\right)\left(x^2-x-1\right)\)
Vậy:
\(f\left(x\right)=0\Rightarrow\left[{}\begin{matrix}x=\frac{-1\pm\sqrt{13}}{2}\\x=\frac{1\pm\sqrt{5}}{2}\end{matrix}\right.\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< \frac{-1-\sqrt{13}}{2}\\\frac{1-\sqrt{5}}{2}< x< \frac{1+\sqrt{5}}{2}\\x>\frac{-1+\sqrt{13}}{2}\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}\frac{-1-\sqrt{13}}{2}< x< \frac{1-\sqrt{5}}{2}\\\frac{1+\sqrt{5}}{2}< x< \frac{-1+\sqrt{13}}{2}\end{matrix}\right.\)
6.
\(f\left(x\right)=\frac{x^2+4x+15-\left(x-3\right)\left(x-1\right)+\left(x-2\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x^2+7x+10}{\left(x-1\right)\left(x+1\right)}=\frac{\left(x+5\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định khi \(x=\pm1\)
\(f\left(x\right)=0\Rightarrow x=\left\{-2;-5\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -5\\-2< x< -1\\x>1\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}-5< x< -2\\-1< x< 1\end{matrix}\right.\)