a, Ta có: f(x)= x2-10x+27 = (x-5)2+2>0
=> pt vô nghiệm
b, g(x)=x2+(2/3)x+4/9=x2+2.(1/3).x+1/9+1/3
= (x+1/3)2+1/3>0
=> pt vô nghiệm.
\(a,f\left(x\right)=x^2-10x+27\)
\(\Rightarrow f\left(x\right)=x^2-5x-5x+25+2\)
\(\Rightarrow f\left(x\right)=x\left(x-5\right)-5\left(x-5\right)+2\)
\(\Rightarrow f\left(x\right)=\left(x-5\right)^2+2\ge2>0\) (Vì \(\left(x-5\right)^2\ge0\) \(Vx\) )
Vậy đa thức f(x) vô nghiệm
\(b,g\left(x\right)=x^2+\frac{2}{3}x+\frac{4}{9}\)
\(\Rightarrow g\left(x\right)=x^2+\frac{1}{3}x+\frac{1}{3}x+\frac{1}{9}+\frac{3}{9}\)
\(\Rightarrow g\left(x\right)=x\left(x+\frac{1}{3}\right)+\frac{1}{3}\left(x+\frac{1}{3}\right)+\frac{1}{3}\)
\(\Rightarrow g\left(x\right)=\left(x+\frac{1}{3}\right)^2+\frac{1}{3}\ge\frac{1}{3}>0\) (Vì \(\left(x+\frac{1}{3}\right)^2\ge0\) \(Vx\) )
Vậy đa thức g(x) vô nghiệm
