a) \(3x^2-5x+2=3x^2-3x-2x+2=3x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(3x-2\right)\)
Để \(3x^2-5x+2>0\Rightarrow\left(x-1\right)\left(3x-2\right)>0\)
Suy ra x - 1 và 3x - 2 đồng dấu. Xét hai trường hợp:
\(\left\{{}\begin{matrix}x-1>0\\3x-2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>1\\x>\frac{2}{3}\end{matrix}\right.\Leftrightarrow x>1\)
TH2; \(\left\{{}\begin{matrix}x-1< 0\\3x-2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 1\\x< \frac{2}{3}\end{matrix}\right.\Leftrightarrow x< \frac{2}{3}\)
b) Tí làm
b) \(x^2-2x+3=\left(x^2-2x+1\right)+2=\left(x-1\right)^2+2\ge2>0\forall x\) nên bất phương trình trên luôn đúng
\(a,3x^2-5x+2>0\)
\(\Leftrightarrow3x^2-6x+x-2>0\)
\(\Leftrightarrow\left(x-2\right)\left(3x+1\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\3x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\3x+1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -\frac{1}{3}\end{matrix}\right.\)
\(b,x^2-2x+3>0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+2>0\)
\(\Leftrightarrow\left(x-1\right)^2+2>0\) (luôn đúng)
