a, Vì \(2x^2\ge0\forall x\)
\(\Rightarrow P\left(x\right)=2x^2+1\ge1>0\)
Vậy P(x) vô nghiệm
b, Vì \(x^4\ge0\forall x;2x^2\ge0\forall x\)
\(\Rightarrow x^4+2x^2\ge0\)
\(\Rightarrow Q\left(x\right)=x^4+2x^2+1\ge1>0\)
Vậy...
c, \(M\left(x\right)=x^2+2x+2018=x^2+x+x+1+2017=x\left(x+1\right)+\left(x+1\right)+2017=\left(x+1\right)\left(x+1\right)+2017=\left(x+1\right)^2+2017\)
Vì \(\left(x+1\right)^2\ge0\forall x\)
\(\Rightarrow M\left(x\right)=\left(x+1\right)^2+2017\ge2017>0\)
Vậy...
d, \(N\left(x\right)=x^2-4x+5=x^2-2x-2x+4+1=x\left(x-2\right)-2\left(x-2\right)+1=\left(x-2\right)\left(x-2\right)+1=\left(x-2\right)^2+1\)
Vì \(\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow N\left(x\right)=\left(x-2\right)^2+1\ge1>0\)
Vậy...
P(x)=2x²+1
Ta có:2x²>;=0 và 1>0
=>P(x)=2x²+1>0
nên đa thức P(x) vô nghiệm
