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

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

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

a) x² - 8x + 19 = ( x² - 8x + 16 ) + 3 = ( x - 4 )² + 3 ≥ 3 > 0 ∀ x ( đpcm )

b) x² + y² - 4x + 2 = ( x² - 4x + 4 ) + y² - 2 = ( x - 2 )² + y² - 2 ≥ -2 ∀ x, y ( chưa cm được -- )

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

d) x² - 2xy + 2y² + 2y + 5 = ( x² - 2xy + y² ) + ( y² + 2y + 1 ) + 4 = ( x - y )² + ( y + 1 )² + 4 ≥ 4 > 0 ∀ x, y ( đpcm )

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

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

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

A=x²-2x+2

A=(x²-2x+1)+1

A=(x-1)²+1

(x-1)²\(\ge\)0 với mọi x

=> (x-1)²+1 >0 hay A>0

Vậy A luôn dương với mọi x,y,z

B=x²+y²+z²+4x-2y-4z+10

B=(x²+4x+4)+(y²-2y+1)+(z²-4z+4)+1

B=(x+2)²+(y-1)²+(z-2)²+1

(x+2)²\(\ge\)0 với mọi x

(y-1)²\(\ge\)0 với mọi y

(z-2)²\(\ge\)0 với mọi z

=>(x+2)²+(y-1)²+(z-2)²+1>0 hay B>0

Vậy B luôn dương với mọi x,y,z

C=x²+y²+2x-4y+6

C=(x²+2x+1)+(y²-4y+4)+1

C=(x+1)²+(y-2)²+1

(x+1)²\(\ge\)0 với mọi x

(y-2)²\(\ge\)0 với mọi y

=>(x+1)²+(y-2)²+1>0 hay C>0

Vậy C luôn dương với mọi x,y,z

a/ \(A=x^2-2x+2\\A=x^2-2x+1+1\\ A=\left(x-1\right)^2+1>0 \)

b/ \(B=x^2+y^2+z^2+4x-2y-4z+10\)

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

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

c/ \(C=x^2+y^2+2x-4y+6\)

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

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

thanks