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

=x^2-2x+1+1

=(x-1)^2+1>0

a: Ta có: \(-x^2+4x-5\)

\(=-\left(x^2-4x+5\right)\)

\(=-\left(x^2-4x+4+1\right)\)

\(=-\left(x-2\right)^2-1< 0\forall x\)

b: Ta có: \(x^4\ge0\forall x\)

\(3x^2\ge0\forall x\)

Do đó: \(x^4+3x^2\ge0\forall x\)

\(\Leftrightarrow x^4+3x^2+3>0\forall x\)

c: Ta có: \(\left(x^2+2x+3\right)=\left(x+1\right)^2+2>0\forall x\)

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

Do đó: \(\left(x^2+2x+3\right)\left(x^2+2x+4\right)>0\forall x\)

\(\Leftrightarrow\left(x^2+2x+3\right)\left(x^2+2x+4\right)+3>0\forall x\)​

b: Ta có: \(x^4\ge0\forall x\)

\(3x^2\ge0\forall x\)

Do đó: \(x^4+3x^2\ge0\forall x\)

\(\Leftrightarrow x^4+3x^2+3>0\forall x\)

c: Ta có: \(\left(x^2+2x+3\right)=\left(x+1\right)^2+2>0\forall x\)

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

Do đó: \(\left(x^2+2x+3\right)\left(x^2+2x+4\right)>0\forall x\)

\(\Leftrightarrow\left(x^2+2x+3\right)\left(x^2+2x+4\right)+3>0\forall x\)

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

b: \(=\left(x-5\right)^2-9y^2\)

\(=\left(x-5-3y\right)\left(x-5+3y\right)\)

Bài 1: 

b: \(=\left(x-5\right)^2-9y^2\)

\(=\left(x-5-3y\right)\left(x-5+3y\right)\)

\(1,\\ a,=3x\left(x-3y\right)\\ b,=\left(x-5\right)^2-9y^2=\left(x-3y-5\right)\left(x+3y-5\right)\\ c,=3x\left(x-y\right)-2\left(x-y\right)=\left(3x-2\right)\left(x-y\right)\\ 2,\\ Sửa:x^2-6x+10=\left(x-3\right)^2+1\ge1>0,\forall x\)

1, =3x (2x -3y)

c, = 3x(x-y) -2(x-y)

= (3x-2)(x-y)

2, Ta có: x² -6x+10= (x-3)² +11

Nhận xét: (x-3)² >= 0 với mọi số thực x

=> (x-3)² +1 >= 1 >0 (đpcm)

x²-6xy+y²+1>0
(x-y)²+1>0
mà (x-y)^2>0
 

\(-25x^2+5x-1=-\left(25x^2-5x+\dfrac{1}{4}\right)-\dfrac{3}{4}=-\left(5x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}\le-\dfrac{3}{4}< 0\forall x\)

a) Đề sai thì phải.Phải là CM: \(x^2-x+1>0\) với mọi x

Ta có:

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

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

Vì \(\left(x-\frac{1}{2}\right)^2\ge0\) nên \(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)

Vậy \(x^2-x+1>0\) với mọi \(x\in R\)

b)Ta có:

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

\(=-\left(x-1\right)^2-3\)

Vì \(-\left(x-1\right)^2\le0\) với mọi x nên \(-\left(x-1\right)^2-3< 0\)

Vậy \(-x^2+2x-4< 0\) với mọi \(x\in R\)