Đặt \(A=9x^2-6x+10\)
Ta có :
\(A\left(x\right)=9x^2-6x+10\)
\(=\left[9x^2-6x+1\right]+9\)
\(=\left[\left(3x^2\right)-3x-3x+1\right]+9\)
\(=\left[3x\left(3x-1\right)-\left(3x-1\right)\right]+9\)
\(=\left(3x-1\right)^2+9\)
Ta có : \(\left(3x-1\right)^2\ge0\)
\(\Rightarrow A\left(x\right)=\left(3x-1\right)^2+9\ge9>0\)
\(\Rightarrow\) Đa thức vô nghiệm