Ta có: \(-x^2+3x-4\)

\(=-\left(x^2-3x+4\right)\)

\(=-\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{7}{4}\right)\)

\(=-\left(x-\dfrac{3}{2}\right)^2-\dfrac{7}{4}< 0\forall x\)

$-x^2+3x-4\\=-x^2+2.x.\dfrac{3}{2}-\dfrac{9}{4}-\dfrac{7}{4}\\=-(x-\dfrac{3}{2})^2-\dfrac{7}{4}<0$

=> ĐPCM

a) x² – x + 1 

=(x² – x + 1/4 )+3/4

=(x-1/2)²+3/4

ta có (x-1/2)²>=0

(x-1/2)²​+3/4>=​+3/4>0

vậy (x-1/2)²​+3/4>0 với mọi số thực x

b)  -x²+2x -4

= -x²+2x -1-3

=-(x²-2x +1)-3

=-(x-2)²​-3

ta có (x-2)²>=0

=>-(x-2)²=<0

=>-(x-2)²​-3=<​-3<0

vậy -(x-2)²​-3<0 với mọi số thực x

a: \(A=x^3-27-x^3+3x^2-3x+1-4\left(x^2-4\right)-x\)

\(=3x^2-4x-26-4x^2+16\)

\(=-x^2-4x-10\)

Ta có:

Ta có:  với mọi số thực x

⇒  với mọi số thực x

⇒  với mọi số thực (ĐPCM)

Ta có:

\(x^2-x+1\)

\(=x^2-2\cdot\dfrac{1}{2}\cdot x+\dfrac{1}{4}+\dfrac{3}{4}\)

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

Mà: \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\) và \(\dfrac{3}{4}>0\)

Nên: \(x^2-x+1>0\)

\(x^2-x+1\)

\(=x^2-\dfrac{1}{2}.x-\dfrac{1}{2}.x+\dfrac{1}{4}+\dfrac{3}{4}\)

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

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

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

\(x^2-x+1=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\\ Mà:\left(x-\dfrac{1}{2}\right)^2>0\forall x\in R\\ Vậy:\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\in R\\ Vậy:x^2-x+1>0\forall x\in R\)

\(a,=\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{19}{4}=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}>0\\ b,=-\left(x^2-5x+\dfrac{25}{4}\right)-\dfrac{7}{4}=-\left(x-\dfrac{5}{2}\right)^2-\dfrac{7}{4}\le-\dfrac{7}{4}< 0\)

a,=(x2+3x+94)+194=(x+32)2+194≥194>0b,=−(x2−5x+254)−74=−(x−52)2−74≤−74<0

Ta có:

x2 – 2xy + y2 + 1

= (x2 – 2xy + y2) + 1

= (x – y)2 + 1.

(x – y)2 ≥ 0 với mọi x, y ∈ R

⇒ x2 – 2xy + y2 + 1 = (x – y)2 + 1 ≥ 0 + 1 = 1 > 0 với mọi x, y ∈ R (ĐPCM).

\(x^2-3x+5=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}>0\) với mọi số thực x