Question

Chung to rang da thuc sau khong co nghiem

B(x)=x^2+x+1

Answer 1

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

Answer 2

\(B\left(x\right)=x^2+x+1\)

        \(=\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}\)

        \(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)

.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 .

Answer 3

\(B\left(x\right)=x^2+x+1\)

\(\Delta=\left(-1\right)^2-4.1.1=-3< 0.PTVN\)

→ Đa thức trên không có nghiệm .

CHÚC BẠN HỌC TỐT !!!

Answer 4

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