a) Ta có :  \(x^2-6x+10\)

\(=\left(x^2-6x+9\right)+1\)

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

b) Ta có :  \(4x-x^2-5\)

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

\(=-\left(x-2\right)^2-1\le-1< 0\forall x\)

Vậy ...

Câu a:

Cm: A = \(x^2+x+1>0\forall x\)

A = \(x^2+2.x\).\(\frac12+\left(\frac12\right)^2+\frac34\)

A = [\(x^2+2x\).\(\frac12\) + \(\left(\frac12\right)^2\)] + \(\frac34\)

A = [\(x+\frac12]^2\) + \(\frac34\)

[\(x+\frac12\)]\(^2\) ≥ 0 ∀ \(x\)

A = [\(x+\frac12\)]\(^2\) + \(\frac34\) ≥ \(\frac34\forall x\)

A > 0 \(\forall x\) (đpcm)

b; B = \(x^{2}\) - \(x + 1\)

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

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

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

Vì [\(x - \frac{1}{2}\)]\(^{2}\) ≥ 0 ∀ \(x\)

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

B > 0 \(\forall x\) (đpcm)

a. \(x^2+3x+5\)

\(=x^2+2.x^2.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{11}{4}\)

\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)

=> đpcm

b. \(4x^2+5x+7\)

\(=\left(2x\right)^2-2.2x.\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{87}{16}\)

= \(\left(2x+\dfrac{5}{4}\right)^2\) + \(\dfrac{87}{16}\) \(\ge\dfrac{87}{16}\)

=> đpcm

b) Ta có: \(2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)

  \(2x^2+2y^2-x^2-2xy-y^2\ge0\) 

\(x^2-2xy+y^2\ge0\)

\(\left(x-y\right)^2\ge0\)  luôn đúng \(\forall x;y\)

Vậy \(2\left(x^2+y^2\right)\ge\left(x+y\right)^2\left(đpcm\right)\)

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

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

c)\(9x^2+12x+10=\left(9x^2+12x+4\right)+6=\left(3x+2\right)^2+6\ge6>0\)

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