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

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

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

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

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

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

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

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 \(x-4=0\Rightarrow x=4\)

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

\(=-\left(x^2+8x-5\right)\)

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

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

Min bằng 21 \(\Leftrightarrow x=-4\)

a) = 3(x²-2x+1) +1-3

GTNN = -2

B) tt

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

sử dụng tính chất: /a/+/b/ >= /a+b/

1) \(f\left(x\right)=-3x^2-12x+5\)

\(\Rightarrow f\left(x\right)=-3\left(x^2+4x\right)+5\)

\(\Rightarrow f\left(x\right)=-3\left(x^2+4x+4\right)+5+12\)

\(\Rightarrow f\left(x\right)=-3\left(x+2\right)^2+17\le17\left(-3\left(x+2\right)^2\le0,\forall x\right)\)

\(\Rightarrow GTLN\left(f\left(x\right)\right)=17\left(tạix=-2\right)\)

2) \(f\left(x\right)=-8x^2+20x\)\

\(\Rightarrow f\left(x\right)=-8\left(x^2+\dfrac{5}{2}x\right)\)

\(\Rightarrow f\left(x\right)=-8\left(x^2+\dfrac{5}{2}x+\dfrac{25}{16}\right)+\dfrac{25}{2}\)

\(\Rightarrow f\left(x\right)=-8\left(x+\dfrac{5}{4}\right)^2+\dfrac{25}{2}\le\dfrac{25}{2}\left(-8\left(x+\dfrac{5}{4}\right)^2\le0,\forall x\right)\)

\(\Rightarrow GTLN\left(f\left(x\right)\right)=\dfrac{25}{2}\left(tạix=-\dfrac{5}{4}\right)\)

1) �(�)=−3�2−12�+5f(x)=−3x2−12x+5

⇒�(�)=−3(�2+4�)+5⇒f(x)=−3(x2+4x)+5

⇒�(�)=−3(�2+4�+4)+5+12⇒f(x)=−3(x2+4x+4)+5+12

⇒�(�)=−3(�+2)2+17≤17(−3(�+2)2≤0,∀�)⇒f(x)=−3(x+2)2+17≤17(−3(x+2)2≤0,∀x)

$\Rightarrow GTLN\left(f\left(x\right)\right)=17\left(tạix=-2\right)$

I zì:vv

a) Ta có: \(A=4x^2+4x+11=4x^2+4x+1=10=\left(2x+1\right)^2+10\ge10\forall x\)

Vậy Min_A=10 khi \(x=-\dfrac{1}{2}\)

b) Ta có: \(B=5-8x-x^2=-\left(x^2+8x-5\right)=-\left(x^2+8x+16-21\right)\)

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

Vậy Max_B=21 khi x=-4

A= 5-8x-x²

=-x²-8x+21-16

=21-(x²+8x+16)

=21-(x+4)²\(\ge\)21-0=21

Dấu = khi x=-4

Vậy Amax=21 khi x=-4

B= x²+x+1

\(=x^2+\frac{x}{2}+\frac{x}{2}+\frac{1}{4}+\frac{3}{4}\)

\(=x\left(x+\frac{1}{2}\right)+\frac{1}{2}\left(x+\frac{1}{2}\right)+\frac{3}{4}\)

\(=\left(x+\frac{1}{2}\right)\left(x+\frac{1}{2}\right)+\frac{3}{4}\)

\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge0+\frac{3}{4}=\frac{3}{4}\)

Dấu = khi x=-1/2

Vậy Bmin=3/4 khi x=-1/2

\(A=5-8x-x^2\) 

\(=-\left(x^2+8x+16\right)+21\) 

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

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~ :)

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

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

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

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