\(A=2x\left(6-x\right)\le\dfrac{1}{2}\left(x+6-x\right)^2=18\)

Dấu "=" xảy ra khi \(x=3\)

\(B^2=x^2\left(9-x\right)=-x^3+9x^2\)

\(B^2=-x^3+9x^2-108+108=108-\left(x-6\right)^2\left(x+3\right)\le108\)

\(\Leftrightarrow B\le6\sqrt{3}\)

\(C^2=\left(6-x\right)^2x=32-\left(8-x\right)\left(x-2\right)^2\le32\)

\(\Rightarrow C\le4\sqrt{2}\)

2A = 2x (12 - 2x)

Áp dụng bất đẳng thức cosi

2x (12 - 2x) ≤ \(\dfrac{\left(2x+12-2x\right)^2}{4}\)

⇔ 2A ≤ 36

⇔ A ≤ 18

Dấu bằng xảy ra khi \(\left\{{}\begin{matrix}0\le x\le6\\2x=12-2x\end{matrix}\right.\)⇔ x = 3

Vậy A_max = 18 khi x = 3

\(A=2x\left(6-x\right)\)

\(=-2x^2+12x+18\)

\(=-2\left(x^2-6x+9\right)+18\)

\(=-2\left(x-3\right)^2+18\le18\)

\(maxA=18\Leftrightarrow x-3=0\Leftrightarrow x=3\)

b/ B = \(x^3+\dfrac{3}{x^2}=\dfrac{x^3}{2}+\dfrac{x^3}{2}+\dfrac{1}{x^2}+\dfrac{1}{x^2}+\dfrac{1}{x^2}\ge5\sqrt[5]{\dfrac{x^3}{2}\cdot\dfrac{x^3}{2}\cdot\dfrac{1}{x^2}\cdot\dfrac{1}{x^2}\cdot\dfrac{1}{x^2}}=5\sqrt[5]{\dfrac{1}{4}}\)

Dấu ''='' xảy ra khi \(\dfrac{x^3}{2}=\dfrac{1}{x^2}\Leftrightarrow x^5=2\Leftrightarrow x=\sqrt[5]{2}\)

Vậy: \(MIN_B=5\sqrt[5]{\dfrac{1}{4}}\Leftrightarrow x=\sqrt[5]{2}\)

Ta có : - 2 ≤ x ≤ 3

⇒ x + 2 ≥ 0 và 3 - x ≥ 0

Áp dụng BĐT Cô - Si , ta có :

a² + b² ≥ 2ab ( a > 0 ; b > 0)

⇔ ( a + b)² ≥ 4ab

⇔\(\dfrac{\left(a+b\right)^2}{4}\)≥ ab

⇒ A = ( x + 2)( 3 - x) ≤ \(\left[\dfrac{\left(x+2\right)+\left(3-x\right)}{2}\right]^2=\left(\dfrac{5}{2}\right)^2=\dfrac{25}{4}\)

⇒ A_MAX = \(\dfrac{25}{4}\) ⇔ x + 2 = 3 - x ⇔ x = \(\dfrac{1}{2}\)

\(Q=x^2\left(4-3x\right)=\dfrac{4}{9}.\dfrac{3}{2}x.\dfrac{3}{2}x\left(4-3x\right)\)

\(Q\le\dfrac{1}{27}.\dfrac{4}{9}.\left(\dfrac{3x}{2}+\dfrac{3x}{2}+4-3x\right)^3=\dfrac{256}{243}\)

\(Q_{maxx}=\dfrac{256}{243}\) khi \(\dfrac{3x}{2}=4-3x\Leftrightarrow x=\dfrac{8}{9}\)

`-2<=x<=2`
`<=>x+2>=0,x-2<=0`
`=>(x+2)(x-2)<=0`
`<=>x^2-4<=0`
`<=>x^2<=4`
`=>A<=4-2x+7=11-2x`
Vì `x>=-2=>2x>=-4`
`=>A<=11+4=15`
Dấu "=" xảy ra khi `x=-2

mng giúp em với ạ

đk câu a hình như sai đấy

Dễ dàng nhận ra \(A\ge0\)

\(A^2=x+3-x+2\sqrt{x\left(3-x\right)}=3+2\sqrt{x\left(3-x\right)}\ge3\)

\(\Rightarrow A\ge\sqrt{3}\)

\(A_{min}=\sqrt{3}\) khi \(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)