a) \(A=x^2+2x+2\)

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

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

b) \(B=4x^2-4x+11\)

\(=4x^2-4x+1+10\)

\(=\left(2x-1\right)^2+10>0\forall x\)

c) \(C=x^2-x+1\)

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

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

d) Ta có: \(D=x^2-2x+y^2+4y+6\)

\(=x^2-2x+1+y^2+4y+4+1\)

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

e) Ta có: \(D=x^2-2xy+y^2+x^2-8x+20\)

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

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

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

\(B=4x^2-4x+11=\left(2x-1\right)^2+10\ge10>0\left(\forall x\right)\)

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

\(D=x^2-2x+y^2+4y+6=x^2-2x+1+y^2+4y+4+1\)

\(=\left(x-1\right)^2+\left(y-2\right)^2+1\ge1>0\)

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

\(=\left(x-y\right)^2+\left(x-4\right)^2+4\ge4>0\)\(\left(\forall x,y\right)\)

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

c) \(C=4x-10-x^2=-\left(x^2-4x+10\right)\)

\(=-\left(x^2-4x+4+6\right)=-\left[\left(x-2\right)^2+6\right]\)

\(=-\left(x^2-4x+4+6\right)=-\left[\left(x-2\right)^2\right]-6\le-6< 0\forall x\)

sai đề rồi x=-2;y=5 biểu thức dương

\(3x-x^2-7=-\left(x^2-3x+7\right)=-\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}\right)-\frac{19}{4}\)

\(=-\left(x-\frac{3}{2}\right)^2-\frac{19}{4}\le-\frac{19}{4}\)với mọi \(x\).

Do đó ta có đpcm.

\(a,P=5x\left(2-x\right)-\left(x+1\right)\left(x+9\right)\)

\(=10x-5x^2-\left(x^2+x+9x+9\right)\)

\(=10x-5x^2-x^2-x-9x-9\)

\(=\left(10x-x-9x\right)+\left(-5x^2-x^2\right)-9\)

\(=-6x^2-9\)

Ta thấy: \(x^2\ge0\forall x\)

\(\Rightarrow-6x^2\le0\forall x\)

\(\Rightarrow-6x^2-9\le-9< 0\forall x\)

hay \(P\) luôn nhận giá trị âm với mọi giá trị của biến \(x\).

\(b,Q=3x^2+x\left(x-4y\right)-2x\left(6-2y\right)+12x+1\)

\(=3x^2+x^2-4xy-12x+4xy+12x+1\)

\(=\left(3x^2+x^2\right)+\left(-4xy+4xy\right)+\left(-12x+12x\right)+1\)

\(=4x^2+1\)

Ta thấy: \(x^2\ge0\forall x\)

\(\Rightarrow4x^2\ge0\forall x\)

\(\Rightarrow4x^2+1\ge1>0\forall x\)

hay \(Q\) luôn nhận giá trị dương với mọi giá trị của biến \(x\) và \(y\).

#\(Toru\)

\(A=x^2+10y^2+2xy-6y+5\)

\(A=x^2+2xy+y^2+9y^2-6y+1+4\)

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

Mà \(\hept{\begin{cases}\left(x+y\right)^2\ge0\\\left(3y+1\right)^2\ge0\\4>0\end{cases}}\)

=> A luôn dương với mọi x ; y

\(B=x-x^2-1\)

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

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

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

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

Mà \(\hept{\begin{cases}-\left(x-\frac{1}{2}\right)^2\le0\\-\frac{3}{4}< 0\end{cases}}\)

=> B luôn âm với mọi x

Ta có A = -x² + 4x - 6 - y² - 2y 

= -(x² - 4x + 4) - (y² + 2y + 1) - 1

= -(x - 2)² - (y + 1)² - 1 \(\le-1< 0\)

=> A < 0 với mọi x ; y

A = -x² + 4x - 6 - y² - 2y 

= -( x² - 4x + 4 ) - ( y² + 2y + 1 ) - 1

= -( x - 2 )² - ( y - 1 )² - 1 ≤ -1 < 0 ∀ x, y

=> đpcm

\(A=-x^2+4x-6-y^2-2y\)

\(=-x^2+4x-4-y^2-2y-1-1\)

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

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

\(=-\left[\left(x-2\right)^2+\left(y+1\right)^2+1\right]\)

mà \(\left(x-2\right)^2\ge0\forall x\)

      \(\left(y+1\right)^2\ge0\forall y\)

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

\(\Rightarrow\left(x-2\right)^2+\left(y+1\right)^2+1\ge1\)

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

\(\Rightarrow-\left[\left(x-2\right)^2+\left(y+1\right)^2+1\right]< 0\)

\(\Rightarrow A< 0\)

Vậy A luôn có giá trị âm với mọi x,y

a Ta có 4x² - 4x + 3 = (4x² - 4x + 1) + 2 = (2x - 1)² + 2 \(\ge\)2 > 0 (đpcm)

b) Ta có y - y² - 1 

= -(y² - y + 1)

= -(y² - y + 1/4) - 3/4

= -(y - 1/2)² - 3/4 \(\le-\frac{3}{4}< 0\)(đpcm)

a) 4x² - 4x + 3 = ( 4x² - 4x + 1 ) + 2 = ( 2x - 1 )² + 2 ≥ 2 > 0 ∀ x ( đpcm )

b) y - y² - 1 = -( y² - y + 1/4 ) - 3/4 = -( y - 1/2 ) - 3/4 ≤ -3/4 < 0 ∀ x ( đpcm )

4x² - 4x + 3 

= 4x² - 4x + 1 + 2

= 4 ( x -\(\frac{1}{2}\))² + 2\(\ge2\)

=> Đpcm

y - y² - 1

= - y² + y -\(\frac{1}{4}-\frac{3}{4}\)

= - ( y -\(\frac{1}{2}\))² -\(\frac{3}{4}\le-\frac{3}{4}\)

=> Đpcm