x2+6x+10
=x2+3x+3x+3.3+1
=x(3+x)+3(3+x)+1
=(3+x)(3+x)+1
=(3+x)2+1
Vì (3+x)2>hoặc=0
=> (3+x)2+1>1
Vậy đa thức trên ko có nghiệm
\(B\left(x\right)=x^2+x+1\)
\(=\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
.Ta có : \(\left(x+\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
\(\Rightarrow B\left(x\right)>0\) với mọi x
Vậy \(B\left(x\right)\) vô nghiệm .
\(B\left(x\right)=x^2+x+1\)
\(\Delta=\left(-1\right)^2-4.1.1=-3< 0.PTVN\)
→ Đa thức trên không có nghiệm .
CHÚC BẠN HỌC TỐT !!!
Ta có: B(x)=x2+x+1
=x2+\(\dfrac{1}{2}x\)+ \(\dfrac{1}{2}x\)+\(\dfrac{1}{4}\)+\(\dfrac{3}{4}\)=
=x(x+\(\dfrac{1}{2}\))+\(\dfrac{1}{2}\)(x+\(\dfrac{1}{2}\))+\(\dfrac{3}{4}\)
=(x+\(\dfrac{1}{2}\))(x+\(\dfrac{1}{2}\))+\(\dfrac{3}{4}\)=(x+\(\dfrac{1}{2}\))2+\(\dfrac{3}{4}\).
Do (x\(+\dfrac{1}{2}\))2\(\ge\)0 (\(\forall\)x)
\(\Rightarrow\)(x+\(\dfrac{1}{2}\))2+\(\dfrac{3}{4}\)\(\ge\)\(\dfrac{3}{4}\) (\(\forall\)x)
hay (x+\(\dfrac{1}{2}\))2+\(\dfrac{3}{4}\)>0 (\(\forall\)x)
Vậy B(x) vô nghiệm