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

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

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

Ta có : C = 4x² + 4y² - 8x + 4y + 427

=> C = (4x² - 8x + 4) + (4y² + 4y + 1) + 422

=> C = (2x - 2)² + (2y + 1)² + 422

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

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

Nên C = (2x - 2)² + (2y + 1)² + 422  \(\ge422\forall x\)

Suy ra : C = (2x - 2)² + (2y + 1)² + 422 \(>0\forall x\)

Vậy C luôn luôn dương (đpcm)

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) \(A=x^2+2x+3=x^2+2x+1+2\)

\(=\left(x+1\right)^2+2\ge2\)

Vậy A luôn dương với mọi x

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

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

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

Vậy B luôn âm với mọi x

a)\(x^2+2x+3=\left(x^2+2x+1\right)+2=\left(x+1\right)^2+2\ge2\)

Vậy x² +2x+3 luôn dương.

b)\(-x^2+4x-5=-\left(x^2-4x+5\right)=-\left(x^2-4x+4+1\right)=-\left[\left(x-2\right)^2+1\right]\le-1\)

Vậy -x² +4x-5 luôn luôn âm.

a.x²+ 2x+ 3

=x²+ 2.x.1+ 1²- 1²+ 3

= (x+1)² -1+3

= (x+1)²+ 2

Ta có: (x+1)² ≥0

           (x+1)²+ 3≥ 3>0

⇒x²+ 2x+ 3>0 mọi x

Vậy x²+ 2x+3>0 mọi x

b. -x²+ 4x- 5

= - (x²- 4x +5)

= - (x²- 2.x.2+ 2²- 2²+ 5)

= - ((x- 2)²- 4+ 5)

= - ((x- 2)²+1)

= -(x- 2)² -1

Ta có: (x-2)² ≥0

         - (x-2)² ≤0

         - (x-2)² +1≤ 1

⇒ -x²+ 4x- 5 <0 mọi x

Vậy -x²+ 4x- 5 <0 mọi x

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

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

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

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

Xem lại đề nha, kết quả vẫn có thể âm mà

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: \(D=x^2-2xy+y^2+x^2-8x+20\)

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

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

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

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

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

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

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

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

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

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

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

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

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