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![](https://rs.olm.vn/images/avt/0.png?1311)
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:
\(f\left(x\right)=0\Rightarrow x=\left\{-6;2\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -6\\x>2\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow-6< x< 2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
28. \(x^2+\frac{9x^2}{\left(x-3\right)^2}=40\) DK: \(x\ne3\)
PT\(\Leftrightarrow\left(x+\frac{3x}{x-3}\right)^2-6\frac{x^2}{x-3}-40=0\)\(\Leftrightarrow\frac{x^4}{\left(x-3\right)^2}-6\frac{x^2}{x-3}-40=0\)
Dat \(\frac{x^2}{x-3}=a\). PTTT \(a^2-6a-40=0\)\(\Leftrightarrow\left(a-10\right)\left(a+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=10\\a=-4\end{matrix}\right.\)
giai tiep
14. \(\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}=1\) DK: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
PT\(\Leftrightarrow\frac{\sqrt{x}-1+\sqrt{x}+1}{x-1}=1\Leftrightarrow2\sqrt{x}=x-1\)\(\Leftrightarrow x-2\sqrt{x}+1=2\Leftrightarrow\left(\sqrt{x}-1\right)^2=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3+2\sqrt{2}\\x=3-2\sqrt{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
\(\frac{x^2+2x+5}{x+4}-\left(x-3\right)\ge0\)
\(\Leftrightarrow\frac{x^2+2x+5-\left(x-3\right)\left(x+4\right)}{x+4}\ge0\)
\(\Leftrightarrow\frac{x+17}{x+4}\ge0\Rightarrow\left[{}\begin{matrix}x>-4\\x\le-12\end{matrix}\right.\)
2.
\(\frac{x^2-3x-1}{2-x}+x>0\)
\(\Leftrightarrow\frac{x^2-3x-1+x\left(2-x\right)}{2-x}>0\)
\(\Leftrightarrow\frac{-x-1}{2-x}>0\Rightarrow\left[{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\)
3.
\(\frac{3x-47}{3x-1}-\frac{4x-47}{2x-1}>0\)
\(\Leftrightarrow\frac{\left(3x-47\right)\left(2x-1\right)-\left(4x-47\right)\left(3x-1\right)}{\left(3x-1\right)\left(2x-1\right)}>0\)
\(\Leftrightarrow\frac{-6x\left(x-8\right)}{\left(3x-1\right)\left(2x-1\right)}>0\Rightarrow\left[{}\begin{matrix}0< x< \frac{1}{3}\\\frac{1}{2}< x< 8\end{matrix}\right.\)
4.
\(\frac{x\left(x+2\right)+9}{x+2}-4\ge0\)
\(\Leftrightarrow\frac{x^2+2x+9-4\left(x+2\right)}{x+2}\ge0\)
\(\Leftrightarrow\frac{x^2-2x+1}{x+2}\ge0\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{x+2}\ge0\Rightarrow x>-2\)
5.
\(\frac{\left(x-1\right)^3\left(x+2\right)^4\left(x+6\right)}{\left(x-7\right)^3\left(x-2\right)^2}\le0\Rightarrow\left[{}\begin{matrix}x\le-6\\1\le x< 2\\2< x< 7\\x=-2\end{matrix}\right.\)
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