BPT \(\Leftrightarrow3x^2-6x+3>0\)
\(\Leftrightarrow3\left(x-1\right)^2>0\)
Có \(\left(x-1\right)^2\ge0\forall x\) . Dấu ''='' xảy ra khi x = 1
=> Để \(3\left(x-1\right)^2>0\) thì \(\left(x-1\right)^2\ne0\Leftrightarrow x\ne1\)
Vậy \(3x^2-5x-x+3>0\) \(\Leftrightarrow x\ne1\)
Ta có:
\(3x^2-5x-x+3>0\)
\(\Leftrightarrow3x^2-6x+3>0\)
\(\Leftrightarrow3x^2-3x-3x+3>0\)
\(\Leftrightarrow\left(x-1\right)^2>0\)
\(\Rightarrow x-1>0\)
\(\Rightarrow x>1\)
3x2-5x-x+3>0
\(\Leftrightarrow\)3x2-6x+3>0
\(\Leftrightarrow\)3(x2-2x+1)>0
\(\Leftrightarrow\)3(x-1)2>0
vì 3>0 nên (x-1)2>0
\(\Rightarrow\)x-1>0 hoặc x-1<0
\(\Rightarrow\)x>1 hoặc x<1
