\(A=-x^2-6x+3\)

\(\Rightarrow A=-\left(x^2+6x\right)+3\)

\(\Rightarrow A=-\left(x^2+6x+9-9\right)+3\)

\(\Rightarrow A=-\left(x^2+6x+9\right)+3+9\)

\(\Rightarrow A=-\left(x+3\right)^2+12\le12\left(-\left(x+3\right)^2\le0,\forall x\right)\)

\(\Rightarrow Max\left(A\right)=12\left(tạix=-3\right)\)

Max(A)=12(tạix=−3)

A=4x-x^2+3

= -x^2+4x+3

= -(x^2-4x-3)

= -(x^2-2*2x*1+1-4)

= -(x-1)^2+4 <4

GTLN của A là 4 khi x=1

Câu B có vấn đề bạn ơi

C=4x-x^2+1

= -x^2+4x+1

= -(x^2-4x-1)

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

= -(x-1)^2+2 < 2

GTLN của C là 2 khi x=1

\(A=\left(x-3\right)^2+\left(x-11\right)^2\)

\(A=x^2-6x+9+x^2-22x+121\)

\(A=2x^2-28x+130\)

\(A=2\left(x^2-14x+49\right)+32\)

\(A=2\left(x-7\right)^2+32\ge32\)

Vậy GTNN của A là 32 khi x = 7

\(A=19-6x-9x^2 \)

\(A=-\left(9x^2+6x+1\right)+20\)

\(A=-\left(3x+1\right)^2+20\le20\)

Vậy GTLN của A là 20 khi x = \(-\frac{1}{3}\)

Ta có : 

\(A=\dfrac{6x+8}{x^2+1}\)

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

\(=\dfrac{\left(x+3\right)^2}{x^2+1}-1\)

Vì \(\left(x+3\right)^2\ge0\) nên \(\dfrac{\left(x+3\right)^2}{x^2+1}\)

nên \(\dfrac{\left(x+3\right)^2}{x^2+1}-1\ge-1\) hay \(A>-1\)

Dấu ' = ' xảy ra khi \(x=-3\)

Vậy \(A_{min}=-1\) khi \(x=-3\)

Ta có :

\(A=\dfrac{6x+8}{x^2+1}\)

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

\(=-\dfrac{\left(3x-1\right)^2}{x+1}+9\)

Vì \(-\dfrac{\left(3x-1\right)^2}{x^2+1}\le0\) nên \(-\dfrac{\left(3x-1\right)^2}{x^2+1}+9\le9\)

Dấu '' = '' xảy ra khi \(x=\dfrac{1}{3}\)

Vậy \(A_{max}=9\) khi \(x=\dfrac{1}{3}\)

\(A=-x^2+6x+2=-\left(x-3\right)^2+11\le11\)

Vậy Max  \(A=11\)khi  \(x=3\)

\(B=-x^2-4x=-\left(x+2\right)^2+4\le4\)

Vậy Max \(B=4\)khi  \(x=-2\)

\(C=-2x^2+6x+3=-2\left(x-\frac{3}{2}\right)^2+\frac{15}{2}\le\frac{15}{2}\)

Vậy Max \(C=\frac{15}{2}\)khi  \(x=\frac{3}{2}\)

Giang sai rồi nhá , nó ko chỉ có max đâu , nó có cả Min nữa đấy

\(A=\dfrac{3x^2-6x+17}{x^2-2x+5}\)

\(=3+\dfrac{2}{x^2-2x+5}\)

Mà \(x^2-2x+5\ge4\)

=> \(\dfrac{2}{x^2-2x+5}\le\dfrac{1}{2}\)

=> A ≤ 7/2

Dấu "=" xảy ra ⇔ x = 1

Ta có : \(A=\dfrac{3x^2-6x+17}{x^2-2x+5}=\dfrac{3x^2-6x+15+2}{x^2-2x+5}=\dfrac{3\left(x^2-2x+5\right)+2}{x^2-2x+5}\)

\(=3+\dfrac{2}{x^2-2x+5}\)

- Thấy : \(x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\)

Lại có : \(\left(x-1\right)^2\ge0\forall x\)

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

\(\Rightarrow\dfrac{2}{x^2-2x+5}\le\dfrac{2}{4}=\dfrac{1}{2}\)

\(\Rightarrow3+\dfrac{2}{x^2-2x+5}\le\dfrac{7}{2}\)

\(HayA\le\dfrac{7}{2}\)

Vậy Max_A = \(\dfrac{7}{2}\) Dấu " = " xảy ra <=> x - 1 = 0

<=> x = 1 .

Bài 1 :

\(A=-x^2+6x+14\)

\(A=-x^2+6x-9+23\)

\(A=-\left(x^2-6x+9\right)+23\)

\(A=-\left(x-3\right)^2+23\)

Vì \(-\left(x-3\right)^2\le0\)

\(\Rightarrow A=-\left(x-3\right)^2+23\le23\)

\(\Rightarrow Max\left(A\right)=23\)

Bài 2 :

\(B=4x^2+12x+30\)

\(\Rightarrow B=4x^2+12x+9+21\)

\(\Rightarrow B=\left(2x+3\right)^2+21\)

Vì \(\left(2x+3\right)^2\ge0\)

\(\Rightarrow B=\left(2x+3\right)^2+21\ge21\)

\(\Rightarrow Min\left(B\right)=21\)