Ta có : D = 5 - 8x - x² 

= -x² - 8x - 16 + 16 + 5 

= -(x² + 8x + 16) + 21

= -(x + 4)² + 21

Mà : -(x + 4)² \(\le0\forall x\) 

Nên : -(x + 4)² + 21 \(\le21\forall x\)

Vậy D_max = 21 , dấu "=" xảy ra khi và chỉ khi x = -4

\(D=5-8x-x^2=-x^2-8x-16+21=-\left(x^2+2.x.4+4^2\right)+21\)

\(=-\left(x+4\right)^2+21\le21\) Có GTNN là 21 tại x = - 4

Vậy ........

-x^2-8x-16+16+5=-(x-4)^2+21 \(\le\)21

Dấu "=" xảy ra khi x=4

2:

a: =-(x^2-12x-20)

=-(x^2-12x+36-56)

=-(x-6)^2+56<=56

Dấu = xảy ra khi x=6

b: =-(x^2+6x-7)

=-(x^2+6x+9-16)

=-(x+3)^2+16<=16

Dấu = xảy ra khi x=-3

c: =-(x^2-x-1)

=-(x^2-x+1/4-5/4)

=-(x-1/2)^2+5/4<=5/4

Dấu = xảy ra khi x=1/2

1) 

a) \(A=x^2+4x+17\)

\(A=x^2+4x+4+13\)

\(A=\left(x+2\right)^2+13\) 

Mà: \(\left(x+2\right)^2\ge0\) nên \(A=\left(x+2\right)^2+13\ge13\)

Dấu "=" xảy ra: \(\left(x+2\right)^2+13=13\Leftrightarrow x=-2\)

Vậy: \(A_{min}=13\) khi \(x=-2\)

b) \(B=x^2-8x+100\)

\(B=x^2-8x+16+84\)

\(B=\left(x-4\right)^2+84\)

Mà: \(\left(x-4\right)^2\ge0\) nên: \(A=\left(x-4\right)^2+84\ge84\)

Dấu "=" xảy ra: \(\left(x-4\right)^2+84=84\Leftrightarrow x=4\)

Vậy: \(B_{min}=84\) khi \(x=4\)

c) \(C=x^2+x+5\)

\(C=x^2+x+\dfrac{1}{4}+\dfrac{19}{4}\)

\(C=\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}\)

Mà: \(\left(x+\dfrac{1}{2}\right)^2\ge0\) nên \(A=\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)

Dấu "=" xảy ra: \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}=\dfrac{19}{4}\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy: \(A_{min}=\dfrac{19}{4}\) khi \(x=-\dfrac{1}{2}\)

1:

a: A=x^2+4x+4+13

=(x+2)^2+13>=13

Dấu = xảy ra khi x=-2

b; =x^2-8x+16+84

=(x-4)^2+84>=84

Dấu = xảy ra khi x=4

c: =x^2+x+1/4+19/4

=(x+1/2)^2+19/4>=19/4

Dấu = xảy ra khi x=-1/2

Đặt  \(P=\frac{x^2-8x+6}{x^2+1}\)

\(\Leftrightarrow P\left(x^2+1\right)=x^2-8x+6\)

\(\Leftrightarrow\left(P-1\right)x^2+8x+\left(P-6\right)=0\)

Ta có  \(\Delta'=16-\left(P-1\right)\left(P-6\right)=-P^2+7P+10\)

Vì  \(\Delta'\ge0\)  \(\Rightarrow-P^2+7P+10\ge0\)

\(\Leftrightarrow\frac{7-\sqrt{89}}{2}\le P\le\frac{7+\sqrt{89}}{2}\)

Vậy GTLN của P là  \(\frac{7+\sqrt{89}}{2}\)

Đặt \(A=\frac{x^2-8x+6}{x^2+1}=1+\frac{5-8x}{x^2+1}\)

Để A max thì 

\(\frac{5-8x}{x^2+1}\) lớn nhất 

Có : \(x^2+1\ge1\)

\(\Rightarrow Max=1\)

<=> x = 0

=> \(\frac{5-8x}{x^2+1}\le\frac{5-8.0}{1}=5\)

Vậy \(Max_A=6\)

<=> x = 0

Kurosaki Akatsu giải sai rồi nha

\(A=2x^2+8x-24\)

\(=2\left(x^2+4x-12\right)\)

\(=2\left[x^2+4x-4-8\right]\)

\(=2\left[\left(x-2\right)^2-8\right]\)

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

\(\Rightarrow\left(x-2\right)^2-8\ge-8\)

\(\Rightarrow2\left[\left(x-2\right)^2-8\right]\ge-16\)

Do đó GTNN của A là -16 khi \(x-2=0\Rightarrow x=2\)

\(B=x^2-8x+5=x^2-8x+16-9\)

\(=x^2-2\left(4x\right)+4^2-9\)

\(=\left(x-4\right)^2-9\)

\(\left(x-4\right)^2\ge0\)

\(\Rightarrow\left(x-4\right)^2-9\ge-9\)

Do đó GTNN của B là -9 khi \(x-4=0\Rightarrow x=4\)

P=(-x^2+8x-7)/(2x+2)

P-1=-(x^2-8x+7+x^2+1)/2(x+1)

P-1=-(2x^2-8x+8)/2(x+1)

P-1=-2(x^2-4x+4)/2(x+1)

P-1=-2(x-2)^2/2(x+1)

Vì -2(x-2)^2/2(x+1) ≥0

=> P-1≥0

=>P≥1

Dấu = xảy ra khi x-2=0 =>x=2

Vậy Pmin = 3 khi x = 2

\(A=5-8x-x^2\) 

\(=-\left(x^2+8x+16\right)+21\) 

\(=-\left(x+4\right)^2+21\le21\forall x\) 

Dấu "=" xảy ra<=> \(-\left(x+4\right)^2=0\Leftrightarrow x=-4\) 

Vậy....

Ta có

=-(x2+ 8x +16) +21

= - (x + 4 ) 2 + 21 < 21x

= - ( x+ 4) 2 = 0<=> = -4

~Study well~ :)

\(A=5-8x-x^2\)

\(=-\left(x2+8x+16\right)+21\)

\(=-\left(x+4\right)^2+21< 21x\)

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

1) \(A=x^2+8x+15=\left(x^2+8x+16\right)-1=\left(x+4\right)^2-1\ge-1\)

\(minA=-1\Leftrightarrow x=-4\)

2) \(B=7x-x^2-5=-\left(x^2-7x+\dfrac{49}{4}\right)+\dfrac{29}{4}=-\left(x-\dfrac{7}{2}\right)^2+\dfrac{29}{4}\le\dfrac{29}{4}\)

\(maxB=\dfrac{29}{4}\Leftrightarrow x=\dfrac{7}{2}\)

Ta có: \(A=x^2+8x+15\)

\(=x^2+8x+16-1\)

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

Dấu '=' xảy ra khi x=-4

Lớp 8 nhé, mình chọn nhầm