Ta có: \(-x^2+4x-5=-\left(x^2-4x+4\right)-1=-\left(x-2\right)^2-1\le-1< 0\forall x\in R\)

`4x-x^2-5`

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

`=-[(x^2-4x+4)+1]`

`=-(x-2)^2-1`

Ta thấy `-(x-2)^2<=0,∀x`

`->-(x-2)^2-1<=-1<0,∀x` ( đpcm )

Ta có: \(-x^2+4x-5\)

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

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

Chả hỉu j cả????

\(-x^2+4x-5=-\left(x^2-4x+5\right)=-\left(x^2-4x-4+1\right)=-\left(x^2-4x-4\right)-1=-\left(x-2\right)^2-1\le-1< 0\)

Ta có: \(x^2-4x+5=\left(x^2-2.x.2+2^2\right)+1\)

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

Vì \(\left(x-2\right)^2\ge0\left(\forall x\right);1\ge0\)

Vậy \(x^2-4x+5\ge0\left(\forall x\right)\)

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

a)  \(A=x^2+2x+5=\left(x+1\right)^2+4\ge4>0\)

Vậy MIN A = 4   khi  x = -1

b)  \(B=x^2+4x+12=\left(x+2\right)^2+8\ge8>0\)

Vậy MIN  B = 8   khi  x = -2

c)  \(C=x^2+6x+31=\left(x+3\right)^2+22\ge22>0\)

Vậy MIN C = 22   khi  x = -3

d) \(D=4x^2+4x+35=\left(2x+1\right)^2+34\ge34>0\)

Vậy MIN  D = 34  khi  x = -1/2

\(A=x^2+2x+5\)

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

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

Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+1\right)^2+4\ge4\forall x\)

\(\Rightarrow A>0\forall x\)

\(A=4\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)

Vậy \(A_{min}=4\Leftrightarrow x=-1\)

\(B=x^2+4x+12\)

\(B=\left(x^2+2.x.2+2^2\right)+8\)

\(B=\left(x+2\right)^2+8\)

đến đó tương tự câu a

\(C=x^2+6x+31\)

\(C=\left(x^2+2.x.3+3^2\right)+22\)

\(C=\left(x+3\right)^2+22\)

đến đó tương tự câu a

\(D=4x^2+4x+35\)

\(D=\left(2x\right)^2+2.2x.1+1+34\)

\(D=\left(2x+1\right)^2+34\)

đến đó tương tự câu a

Tham khảo nhé~

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

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

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