\(x\left(3x^2-x-10\right)>0\)
\(3\left(x^2-2\times x\times\frac{1}{6}+\frac{1}{36}-\frac{1}{36}-10\right)>0\)
\(\left(x-\frac{1}{6}\right)^2>\frac{359}{36}\)
\(\left|x-\frac{1}{6}\right|>\frac{\sqrt{359}}{6}\)
*Th1: \(x>\frac{\sqrt{359}+1}{6}\)
*Th2: \(x< -\frac{\sqrt{359}+1}{6}\)
\(3x^3-x^2-10x>0\)
=> \(x\left(3x^2-x-10\right)>0\)
=> \(x\left(x-2\right)\left(x+\frac{5}{3}\right)>0\)
Ta có TH1 :
\(\hept{\begin{cases}x>0\\x-2>0\\x+\frac{5}{3}>0\end{cases}}=>\hept{\begin{cases}x>0\\x>2\\x>\frac{-5}{3}\end{cases}}\)
TH2 :
\(\hept{\begin{cases}x>0\\x-2< 0\\x+\frac{5}{3}< 0\end{cases}}=>\hept{\begin{cases}x>0\\x< 2\\x< \frac{-5}{3}\end{cases}}\)
TH3 :
\(\hept{\begin{cases}x< 0\\x-2< 0\\x+\frac{5}{3}>0\end{cases}=>\hept{\begin{cases}x< 0\\x< 2\\x>\frac{-5}{3}\end{cases}}}\)
TH4 :
\(\hept{\begin{cases}x< 0\\x-2>0\\x+\frac{5}{3}< 0\end{cases}=>\hept{\begin{cases}x< 0\\x>2\\x< \frac{-5}{3}\end{cases}}}\)
Dài thiệt đó bạn
\(x\left(3x^2-x-10\right)>0\)
\(3\left(x^2-2\times x\times\frac{1}{6}+\frac{1}{36}-\frac{1}{36}-10\right)>0\)
\(\left(x-\frac{1}{6}\right)^2>\frac{359}{36}\)
\(\left|x-\frac{1}{6}\right|>\frac{\sqrt{359}}{6}\)
