\(P\left(x\right)=2x^2+2x+\dfrac{5}{4}\)
\(\Leftrightarrow P\left(x\right)=2\left(x^2+x+\dfrac{5}{16}\right)\)
\(\Leftrightarrow P\left(x\right)=2\left[x^2+\dfrac{1}{2}x+\dfrac{1}{2}x+\dfrac{5}{4}\right]\)
\(\Leftrightarrow P\left(x\right)=2\left[\left(x^2+\dfrac{1}{2}x\right)+\left(\dfrac{1}{2}x+\dfrac{5}{4}\right)\right]\)
\(\Leftrightarrow P\left(x\right)=2\left[x\left(x+\dfrac{1}{2}\right)+\dfrac{1}{2}\left(x+\dfrac{1}{2}\right)+\dfrac{3}{4}\right]\)
\(\Leftrightarrow P\left(x\right)=2\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]\)
Với mọi x ta có :
\(+,\left(x+\dfrac{1}{2}\right)^2\ge0\)
+, \(\dfrac{3}{4}>0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
\(\Leftrightarrow2\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]>0\)
\(\Leftrightarrow P\left(x\right)>0\)
\(\Leftrightarrow P\left(x\right)\) vô nghiệm
P(x)=\(2x^2+2x+\dfrac{5}{4}=2\left(x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}\right)+\dfrac{3}{4}\)
=>\(2\left(x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}\right)=\dfrac{-3}{4}=>2\left(x+\dfrac{1}{2}\right)^2=\dfrac{-3}{4}\left(vônghiemej\right)\)
