\(a.x^2+8x+20=x^2+8x+16+4=\left(x+4\right)^2+4>0\) \(\text{∀}xy\)
\(b.\left(x-7\right)\left(x-11\right)+5\)
Đặt : \(x-9=t,tacó:\left(t+2\right)\left(t-2\right)+5=t^2+1>0\) ∀t
⇒ \(\left(x-7\right)\left(x-11\right)+5>0\) \(\text{∀}x\)
\(c.\left(x^2+1\right)^2-\left(x-1\right)\left(x^2+1\right)\left(x+1\right)=\left(x^2+1\right)\left(x^2+1-x^2+1\right)=2\left(x^2+1\right)>0\) \(\text{∀}x\)
\(d.x^2+x+1=x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+1-\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\text{∀}x\)