\(f\left(x\right)=2x^2+x+1=2\left(x^2+\frac{1}{2}x\right)+1\)
\(=2\left(x^2+2\cdot\frac{1}{4}x+\left(\frac{1}{4}\right)^2-\left(\frac{1}{4}\right)^2\right)+1\)
\(=2\left(x+\frac{1}{4}\right)^2-2\cdot\left(\frac{1}{4}\right)^2+1=2\left(x+\frac{1}{4}\right)^2+\frac{7}{8}\)
Vì \(2\left(x+\frac{1}{4}\right)^2\ge0\) => \(f\left(x\right)=2\left(x+\frac{1}{4}\right)^2+\frac{7}{8}\ge\frac{7}{8}>0\)
=> f(x) vô nghiệm
\(f\left(x\right)=2x^2+x+1=\) 0
( a= 2 ; b = 1 ; c= 1 )
\(\Delta=b^2-4ac\)
\(\Delta=1^2-4.2.1\)
\(\Delta=1-8\)
\(\Delta=-7< 0\)
Vay : phương trình vô nghiệm ( Vi \(\Delta=-7< 0\))
