ta có:

2x²-5x+8=(x²-4x+4)+(x²-x+1/4)+15/4=(x-2)²+(x-1/2)²+15/4>0

a. A= x²-7x+20 = x²-2*\(\dfrac{7}{2}x+\dfrac{49}{4}+\dfrac{31}{4}\)=(x-\(\dfrac{7}{2}\))²+\(\dfrac{31}{4}\)>0 \(\forall x\)(đpcm)

b. B= 2x²+5x+14=2(x²+2*\(\dfrac{5}{4}x+\dfrac{25}{16}+\dfrac{87}{16}\))=2(x+\(\dfrac{5}{4}\))²+\(\dfrac{87}{8}\)>0(đpcm)

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

c: \(A=x^2-6x+9+2=\left(x-3\right)^2+2\ge2\forall x\)

Dấu '=' xảy ra khi x=3

d: \(B=-\left(x^2-4x+5\right)=-\left(x^2-4x+4+1\right)=-\left(x-2\right)^2-1\le-1\forall x\)

Dấu '=' xảy ra khi x=2

Bài 1: \(A=x^2-2x+3\)

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

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

Đẳng thức xảy ra khi \(\left(x-1\right)^2=0\Rightarrow x=1\)

Bài 2:

\(2x^2+4x+11=2x^2+4x+2+9\)

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

\(=2\left(x+1\right)^2+9\ge9>0\forall x\)

a) Ta có: \(B=\dfrac{x^2}{5x+25}+\dfrac{2\left(x+5\right)}{x}+\dfrac{50+5x}{x\left(x+5\right)}\)

\(=\dfrac{x^2}{5\left(x+5\right)}+\dfrac{2\left(x+5\right)}{x}+\dfrac{50+5x}{x\left(x+5\right)}\)

\(=\dfrac{x^3}{5x\left(x+5\right)}+\dfrac{10\left(x+5\right)^2}{5x\left(x+5\right)}+\dfrac{250+25x}{5x\left(x+5\right)}\)

\(=\dfrac{x^3+10x^2+100x+250+250+25x}{5x\left(x+5\right)}\)

\(=\dfrac{x^3+10x^2+125x+500}{5x\left(x+5\right)}\)

\(=\dfrac{x^3+5x^2+5x^2+25x+100x+500}{5x\left(x+5\right)}\)

\(=\dfrac{x^2\left(x+5\right)+5x\left(x+5\right)+100\left(x+5\right)}{5x\left(x+5\right)}\)

\(=\dfrac{\left(x+5\right)\left(x^2+5x+100\right)}{5x\left(x+5\right)}\)

\(=\dfrac{x^2+5x+100}{5x}\)

b) Thay x=-2 vào biểu thức \(B=\dfrac{x^2+5x+100}{5x}\), ta được:

\(B=\dfrac{\left(-2\right)^2+5\cdot\left(-2\right)+100}{-5\cdot2}=\dfrac{4+100-10}{-10}=\dfrac{94}{-10}=-\dfrac{94}{10}=\dfrac{-47}{5}\)

Vậy: Khi x=-2 thì \(B=-\dfrac{47}{5}\)

a) \(x^2-5x+8=\left(x^2-5x+6,25\right)+1,75=\left(x-2,5\right)^2+1,75\ge1,75>0\rightarrowđpcm\)

b) \(-4x^2-4x-2=-\left(4x^2+4x+1\right)-1=-\left(2x+1\right)^2-1\le-1< 0\rightarrowđpcm\)

A =x² -5x +8 >0 với mọi x

= x²-5x+\(\dfrac{25}{4}+\dfrac{7}{4}\)

=\(\left(x-\dfrac{5}{2}\right)^2+\dfrac{7}{4}\)

do \(\left(x-\dfrac{5}{2}\right)^2\ge0\forall x\)

=> \(\left(x-\dfrac{5}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)

=> A luôn lớn hơn 0 vs mọi x

B= -4x² -4x-2 < 0 với mọi x

=-(4x²+4x+2)

=-4x²-4x-1-1

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

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

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

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

do \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\)

=> -4 \(\left(x+\dfrac{1}{2}\right)^2\le0\)

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

vậy B luôn nhỏ hơn 0 vs mọi x