Giải:
\(\left(2x-1\right)\left(3x+1\right)+3x+4\left(3-2x\right)=5\)
\(\Leftrightarrow6x^2-3x+2x-1+3x+12-8x=5\)
\(\Leftrightarrow6x^2+11-6x=5\)
\(\Leftrightarrow6x^2+6-6x=0\)
\(\Leftrightarrow6\left(x^2+1-x\right)=0\)
\(\Leftrightarrow x^2+1-x=0\)
Vì \(x^2+1-x=x^2-x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
\(\Leftrightarrow x^2+1-x\ne0\)
Vậy ...
\(\left(2x-1\right)\left(3x+1\right)+3x+4\left(3-2x\right)=5\)
\(\Leftrightarrow6x^2+2x-3x-1+3x+12-8x=5\)
\(\Leftrightarrow6x^2-6x+6=0\)
\(\Leftrightarrow6\left(x^2-x+1\right)=0\) (1)
Ta co : \(x^2-x+1\)
\(=\left(x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right)-\dfrac{1}{4}+1\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\) \(\ge\dfrac{3}{4}>0\forall x\) (2)
Từ (1)(2) => pt vô nghiệm
