a) Ta có: 9 c 2 – 6c + 3 = ( 3 c   –   1 ) 2 + 2 > 0 "m.

b) Tương tự.

a) \(9c^2-6c+3\)

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

b) \(14m-6m^2-13\)

\(=-6.\left(m^2-\frac{7}{3}m+\frac{13}{6}\right)\)

\(=-6.\left(m^2-2\cdot\frac{7}{6}\cdot m+\frac{49}{36}+\frac{29}{36}\right)\)

\(=-6.\left(m-\frac{7}{6}\right)^2-\frac{29}{6}< 0\)

c) \(a^2-2a+2=\left(a-1\right)^2+1>0\)

d) \(6b-b^2-10=-\left(b^2-6b+9\right)-1=-\left(b-3\right)^2-1< 0\)

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

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) x²-6x+10

=x²-6x+9+1

=(x-3)²+1 \(\ge\) 0 (vì (x-3)²\(\ge\)0)

vậy  x^2-6x+10 luôn luôn dương với mọi x

4x-x²-5

=-x²+4x-4-1

=-(x²-4x+4)-1

=-(x-2)²-1\(\le\)-1 ( vì -(x-2)²\(\le\)0 )

vậy 4x-x^2-5 luôn luôn âm với mọi x

A=x^2+x+1 luon luon dương với mọi x

a)x^2+2x+3

=x^2+2.x.1+1^2+2

=(x+1)^2+2

         Vì (x+1)^2≥0

   Suy ra:(x+1)^2+2≥(đpcm)

b)-x^2+4x-5

=-(x^2-4x+5)

=-(x^2-2.2x+4)-1

=-(x-2)^2-1

             Vì -(x-2)^2≤0

     Suy ra -(x-2)^2-1≤-1(đpcm)

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

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

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

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

`A=x(x-6)+10=x^2-6x+10`

`=x^2 -2.x .3 + 3^2 + 1`

`=(x-3)^2+1 >0 forall x`

`B=x^2-2x+9y^2-6y+3`

`=(x^2-2x+1)+(9y^2-6y+1)+1`

`=(x-1)^2+(3y-1)^2+1 > 0 forall x,y`.