\(a)x^2-2x+y^2+4y+6\\=(x^2-2x+1)+(y^2+4y+4)+1\\=(x^2-2\cdot x\cdot1+1^2)+(y^2+2\cdot y\cdot2+2^2)+1\\=(x-1)^2+(y+2)^2+1\)

Ta thấy: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)

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

hay giá trị của biểu thức trên luôn dương

\(b)x^2-2x+2\\=(x^2-2x+1)+1\\=(x^2-2\cdot x\cdot1+1^2)+1\\=(x-1)^2+1\)

Ta thấy: \(\left(x-1\right)^2\ge0\forall x\)

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

hay giá trị của biểu thức trên luôn dương

A=x²-6x+10

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

\(\Rightarrow A\) luôn dương

A = x² - 6x + 10

= ( x² - 6x + 9 ) + 1 

= ( x - 3 )² + 1 ≥ 1 > 0 ∀ x ( đpcm )

B = x² + x + 5

= ( x² + x + 1/4 ) + 19/4

= ( x + 1/2 )² + 19/4 ≥ 19/4 > 0 ∀ x ( đpcm )

C = 4x² + 4x + 2 

= 4( x² + x + 1/4 ) + 1

= 4( x + 1/2 )² + 1 ≥ 1 > 0 ∀ x ( đpcm )

D = ( x - 3 )( x - 5 ) + 4

= x² - 8x + 15 + 4

= ( x² - 8x + 16 ) + 3 

= ( x - 4 )² + 3 ≥ 3 > 0 ∀ x ( đpcm )

E = x² - 2xy + 1 + y²

= ( x² - 2xy + y² ) + 1 

= ( x - y )² + 1 ≥ 1 > 0 ∀ x, y ( đpcm )

a) \(x^2-3x+8=\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{23}{4}=\left(x-\dfrac{3}{2}\right)^2+\dfrac{23}{4}\ge\dfrac{23}{4}>0\)

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

a: Ta có: \(A=x^2-3x+8\)

\(=x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{23}{4}\)

\(=\left(x-\dfrac{3}{2}\right)^2+\dfrac{23}{4}>0\forall x\)

b: Ta có: \(B=2x^2-2x+2\)

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

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

a, \(A=-x^2+2x-3=-\left(x^2-2x+1-1\right)-3=-\left(x-1\right)^2-2\le-2< 0\forall x\)

Vậy ta có đpcm 

b, \(C=-x^2+4x-7=-\left(x^2-4x+4-4\right)-7=-\left(x-2\right)^2-3\le-3< 0\forall x\)

Vậy ta có đpcm 

c, \(D=-2x^2-6x-5=-2\left(x^2+\frac{2.3}{2}x+\frac{9}{4}-\frac{9}{4}\right)-5\)

\(=-2\left(x+\frac{3}{2}\right)^2-\frac{1}{2}\le-\frac{1}{2}< 0\forall x\)

Vậy ta có đpcm 

d, \(E=-3x^2+4x-4=-3\left(x^2-\frac{4}{3}x+\frac{4}{9}-\frac{4}{9}\right)-4\)

\(=-3\left(x-\frac{2}{3}\right)^2-\frac{8}{3}\le-\frac{8}{3}< 0\forall x\)

Vậy ta có đpcm 

e, tự làm nhé 

a ) \(x^2+6x+10\)

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

\(=\left(x+3\right)^2+1\ge1>0\) ( đpcm )

b ) \(x^2-x+1\)

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

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

x² + 6x + 10

= x² + 2 . x . 3 + 9 + 1

= (x + 3)² + 1

(x + 3)² lớn hơn hoặc bằng 0

(x + 3)² + 1 lớn hơn hoặc bằng 1 > 0 (đpcm)

x² - x + 1

= x² - 2 . x . 1/2 + 1/4 + 3/4

= (x - 1/2)² + 3/4

(x - 1/2)² lớn hơn hoặc bằng 0

(x - 1/2)² + 3/4 lớn hơn hoặc bằng 3/4 > 0 (đpcm)

a/ \(=\left(x^2+2\times x\times3+3^2\right)+1\)

     \(=\left(x+3\right)^2+1\)

  Ta thấy \(\left(x+2\right)^2\ge0\)

 Do đó \(\left(x+3\right)^2+1\ge1>0\)

 Vậy \(x^2+6x+10\) luôn luôn dương

m đâu ????

\(1,\\ A=\left(4x^2+y^2\right)\left(4x^2-y^2\right)=16x^4-y^4\)

Đề sai, biểu thức A ko có m thì sao chứng minh?

\(2,\) Gọi 2 số nguyên lt là \(a;a+1\left(a\in Z\right)\)

Ta có \(a+1-a=1\) là số lẻ (đpcm)

\(3,P=9x^2+24x+16-10x-x^2+16=8x^2+14x+32\)

\(4,Q=x^2-4x+5=\left(x^2-4x+4\right)+1=\left(x-2\right)^2+1\ge1\)

Dấu \("="\Leftrightarrow x-2=0\Leftrightarrow x=2\)

1) m ??

2) Gọi 2 số nguyên liên tiếp là \(a,a+1\left(a\in Z\right)\)

 \(\left(a+1\right)-a=a+1-a=1\) là một số lẻ

3) \(P=\left(3x+4\right)^2-10x-\left(x-4\right)\left(x+4\right)=9x^2+24x+16-10x-x^2+16=8x^2+14x+32\)

4) \(Q=x^2-4x+5=\left(x^2-4x+4\right)+1=\left(x-2\right)^2+1\ge1\)

\(minQ=1\Leftrightarrow x=2\)

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

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

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

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

tự làm tiếp đi chị

a)

\(x^2-4x+9=x^2-4x+4+5=\left(x-2\right)^2+5>0\)

b)

\(4x^2+4x+2017=4\left(x^2+x\right)+2017=4\left(x+\frac{1}{2}\right)^2-1+2017=4\left(x+\frac{1}{2}\right)^2+2016>0\)

c)

\(10-6x+x^2=x^2-6x+10=\left(x-3\right)^2-9+10=\left(x-3\right)^2+1>0\)

d)

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

Bài 1

\(A=x^2-6x+15=x^2-2.3.x+9+6=\left(x-3\right)^2+6>0\forall x\)

\(B=4x^2+4x+7=\left(2x\right)^2+2.2.x+1+6=\left(2x+1\right)^2+6>0\forall x\)

Bài 2

\(A=-9x^2+6x-2021=-\left(9x^2-6x+2021\right)=-\left[\left(3x-1\right)^2+2020\right]=-\left(3x-1\right)^2-2020< 0\forall x\)