Question

Giải bất phương trình sau:

3x²-5x-x+3>0

Answer 1

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 \(3x^2-5x-x+3>0\) \(\Leftrightarrow x\ne1\)

Answer 2

Ta có:

\(3x^2-5x-x+3>0\)

\(\Leftrightarrow3x^2-6x+3>0\)

\(\Leftrightarrow3x^2-3x-3x+3>0\)

\(\Leftrightarrow\left(x-1\right)^2>0\)

\(\Rightarrow x-1>0\)

\(\Rightarrow x>1\)

Answer 3

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