a: Ta có: \(x^2-8x+20\)

\(=x^2-8x+16+4\)

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

b: Ta có: \(-x^2+6x-19\)

\(=-\left(x^2-6x+19\right)\)

\(=-\left(x^2-6x+9+10\right)\)

\(=-\left(x-3\right)^2-10< 0\forall x\)

\(9x^2-6x+2=9x^2-6x+1+1=\left(3x-1\right)^2+1>0\Rightarrowđpcm\)

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

\(25x^2-20x+7=25x^2-20x+4+3=\left(5x-2\right)^2+3>0\left(đpcm\right)\)

\(9x^2-6xy+2y^2+1=\left(9x^2+6xy+y^2\right)+y^2+1=\left(3x+y\right)^2+y^2+1>0\left(đpcm\right)\)

\(\Leftrightarrow x^2+y^2\ge xy;x^2+y^2\ge2\sqrt{x^2y^2}=2\left|xy\right|\ge\left|xy\right|\ge xy\Rightarrowđpcm\)

Cách khác câu e:

\(x^2-xy+y^2=x^2-2x.\frac{y}{2}+\frac{y^2}{4}+\frac{3y^2}{4}=\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}\ge0\forall xy\) (đpcm)

a) 

Đặt \(A=9x^2-6x+2\)

\(=\left(3x\right)^2-2.3x+1+1\)

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

Ta có: \(\left(3x+1\right)^2\ge0;\forall x\)

\(\Rightarrow\left(3x+1\right)^2+1\ge0+1;\forall x\)

Hay \(A\ge1>0;\forall x\)

Các phần khác tương tự cứ việc biến đổi thành hằng đẳng thức

\(a,9x^2-6x+2\)

\(=\left(3x\right)^2-2.3x.1+1^2+1\)

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

Vì\(\left(3x-1\right)^2\ge0\forall x\)

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

\(\Rightarrow9x^2-6x+2>0\forall x\)

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

Vì\(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)

\(\Rightarrow x^2+x+1>0\forall x\)

À xin lỗi sửa sai chút là \(\left(3x-1\right)^2\)nhé 

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

b: \(x-2\cdot\sqrt{x}\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

c: \(=x^2-2\cdot x\cdot\dfrac{1}{2}y+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2=\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2>0\forall x,y\ne0\)

Ta có : \(x^2+2y^2+2xy+y+1\)

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

\(=\left(x+y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x,y\)

a, Sửa đề:

-x²-2x-2

=-(x²+2x+2)

=-(x²+2x+1+1)

=-[(x+1)²+1]<0\(\forall\)x

b, -x²-6x-11

=-(x²+6x+11)

=-(x²+2.x.3+3²+2)

=-[(x+3)²+2]<0\(\forall\)x

Đúng tick nha,

a, -x - 2x - 2

= -(x+2x+1)-1

= -(x+1)² -1

Có (x + 1)² ≥0 ⇒- (x + 1) ≤ 0 ⇒ -(x + 1)² - 1≤ -1

Do đó - x - 2x - 2 < 0 ∀ x

b, -x² - 6x - 11

= -(x² + 2.3.x+ 3²)-2

= -(x+3)² - 2

Có (x + 3)² ≥0 ⇒- (x + 3) ≤ 0 ⇒ -(x + 3)² - 2 ≤ -2

Do đó -x² - 6x - 11 <0 ∀ x