\(A=\left(x-1\right)\left(x-3\right)+2=x^2-4x+3+2=\left(x^2-4x+4\right)+1=\left(x-2\right)^2+1\ge1>0\forall x\)

\(A=x^2+x+1\)

\(A=x^2+x+\dfrac{1}{4}-\dfrac{1}{4}+1\)

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

mà \(\left(x+\dfrac{1}{2}\right)^2\ge0\)

\(\Rightarrow A=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>\dfrac{3}{4}>0\) với mọi x

\(\Rightarrow Dpcm\)

a) \(A=x^2-2x+2=\left(x-1\right)^2+1>0\forall x\inℝ\)

b) \(x-x^2-3=-\left(x^2-x+3\right)\)

\(=-\left(x^2-x+\frac{1}{4}+\frac{11}{4}\right)\)

\(=-\left[\left(x-\frac{1}{2}\right)^2+\frac{11}{4}\right]\)

\(=-\left[\left(x-\frac{1}{2}\right)^2\right]-\frac{11}{4}\le\frac{-11}{4}< 0\forall x\inℝ\)

x²-2x+2=(x²-2x+1)+1=( x-1)²+1

Mà (x-1)²≥0 với mọi x

=> (x-1)²+1>0 với mọi x

=> x²-2x+2>0 với mọi x

x²+x+1=x²+2.x.1/2+1/4+3/4

=(x+1/2)²+3/4

Vì (x+1/2)²\(\ge\)0 nên

(x+1/2)²+3/4>0

=>x²+x+1>0

x2+x+1=x2+2.x.1/2+1/4+3/4 =(x+1/2)2+3/4 Vì (x+1/2)2 ≥ 0 nên (x+1/2)2+3/4>0 =>x2+x+1>0 

Câu a :

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

Vậy biểu thức trên luôn lớn hơn 0 với mọi x

Làm Full cho you nhé,bạn kia sai r:

\(linh_1=x^2+x+1=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\left(đpcm\right)\)

\(linh_2=-4x^2-4x-2=-1\left(4x^2+4x+2\right)=-1\left(4x^2+4x+1+1\right)=-1\left(4x^2+4x+1\right)-1=-1\left(2x+1\right)^2-1< 0\left(đpcm\right)\)