ta có : \(\dfrac{x+4}{x^2-9}-\dfrac{2}{x+3}< \dfrac{4x}{3x-x^2}\)
\(\Leftrightarrow\dfrac{x+4}{x^2-9}-\dfrac{2}{x+3}-\dfrac{4x}{3x-x^2}< 0\)
\(\Leftrightarrow\dfrac{x+4}{\left(x+3\right)\left(x-3\right)}-\dfrac{2}{x+3}-\dfrac{4x}{x\left(3-x\right)}< 0\)
\(\Leftrightarrow\dfrac{x+4}{\left(x+3\right)\left(x-3\right)}-\dfrac{2}{x+3}+\dfrac{4}{x-3}< 0\)
\(\Leftrightarrow\dfrac{x+4-2\left(x-3\right)+4\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}< 0\)
\(\Leftrightarrow\dfrac{x+4-2x+6+4x+12}{\left(x+3\right)\left(x-3\right)}< 0\) \(\Leftrightarrow\dfrac{3x+22}{\left(x+3\right)\left(x-3\right)}< 0\)
ta có : \(3x+22=0\Leftrightarrow x=\dfrac{-22}{3}\)
\(x+3=0\Leftrightarrow x=-3\)
\(x-3=0\Leftrightarrow x=3\)
\(\Rightarrow\) BXD :
| \(x\) | \(-\infty\) | \(\dfrac{-22}{3}\) | \(-3\) | \(3\) | \(+\infty\) | ||||
| \(3x+22\) | \(-\) | \(-\) | \(0\) | \(+\) | \(16\) | \(+\) | \(28\) | \(+\) | \(+\) |
| \(x+3\) | \(-\) | \(-\) | \(\dfrac{-13}{3}\) | \(-\) | \(0\) | \(+\) | \(6\) | \(+\) | \(+\) |
| \(x-3\) | \(-\) | \(-\) | \(\dfrac{-31}{3}\) | \(-\) | \(-6\) | \(-\) | \(0\) | \(+\) | \(+\) |
| \(\dfrac{3x+22}{\left(x+3\right)\left(x-3\right)}\) | \(-\) | \(-\) | \(0\) | \(+\) | oxd | \(-\) | oxđ | \(+\) | \(+\) |
\(\Rightarrow S=\left(-\infty;\dfrac{-22}{3}\right)\cup\left(-3;3\right)\)
vậy ...........................................................................................................