Xét đa thức: Q(x)=2x2-2x+10
Có: 2x2 >= 0
2x < 2x2
=> 2x2- 2x >= 0
Mà 10 >0
=> 2x2-2x+10 >= 10
Vậy đa thức Q(x) vô nghiệm.
Cho x2-2x+10=0
=>x2-2.x.1+12+9=0
=>(x-1)2+9=0 (vô lí vì VT>VP)
=> Q(x) vô nghiệm
Q(x)=2x2-2x+10=2(x2-x+5)=2(x2-x+1+4)
\(Q\left(x\right)=2\left(x^2-\frac{1}{2}x-\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}+4\right)\)
\(Q\left(x\right)=2\left[x\left(x-\frac{1}{2}\right)-\frac{1}{2}\left(x-\frac{1}{2}\right)+\frac{3}{4}+4\right]\)
\(Q\left(x\right)=2\left(x-\frac{1}{2}\right)\left(x-\frac{1}{2}\right)+\frac{3}{4}+4=2\left(x-\frac{1}{2}\right)^2+\frac{19}{4}\)
Vì \(2\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow2\left(x-\frac{1}{2}\right)^2+\frac{19}{4}\ge\frac{19}{4}>0\)
=>Q(x) vô nghiệm
