a) \(A=\dfrac{7}{3}\left(x^2+1\right)\)

Ta có:

\(x^2\ge0\forall x\\ \Rightarrow x^2+1\ge1\forall x\)

Để \(A=\dfrac{7}{3}\left(x^2+1\right)\) đạt GTNN thì \(x^2+1\) đạt GTNN

\(hay:x^2+1=1\)

Thay \(x^2+1=1\) vào \(A=\dfrac{7}{3}\left(x^2+1\right)\) ta có:

\(A=\dfrac{7}{3}.1\\ A=\dfrac{7}{3}\)

Vậy \(Max_A=\dfrac{7}{3}\) tại \(x=0\)

a/ \(A=\dfrac{7}{3}\left(x^2+1\right)\)

Ta có: \(x^2\ge0\forall x\)

\(\Rightarrow\left(x^2+1\right)\ge1\)

Dấu ''='' xảy ra \(\Leftrightarrow x=0\)

\(\Rightarrow Min_A=\dfrac{7}{3}\Leftrightarrow x=0\)

Vậy.................................................

b/ Để \(\dfrac{72-12x}{5-x}\) lớn nhất

\(\Rightarrow5-x\) nhỏ nhất

\(\Rightarrow...................\) tự lm nốt mk fai ik hok r`

nbbbbbnbnbb

Max = vô cùng

Min = 5 (theo mình là vậy)

E= (4x^2-12x+9/4)+(y^2+2xy+x^2)-81/4
=(2x-3/2)^2+(y+x)^2-81/4
Max E= -81/4<=> x=3/4 va` y=-3/4

Không có max

`a)sqrt{x^2-2x+5}`

`=sqrt{x^2-2x+1+4}`

`=sqrt{(x-1)^2+4}`

Vì `(x-1)^2>=0`

`=>(x-1)^2+4>=4`

`=>sqrt{(x-1)^2+4}>=sqrt4=2`

Dấu "=" xảy ra khi `x=1.`

`b)2+sqrt{x^2-4x+5}`

`=2+sqrt{x^2-4x+4+1}`

`=2+sqrt{(x-2)^2+1}`

Vì `(x-2)^2>=0`

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

`=>sqrt{(x-2)^2+1}>=1`

`=>sqrt{(x-2)^2+1}+2>=3`

Dấu "=" xảy ra khi `x=2`

\(A=x^2-2x+50\)

\(A=x^2-2x+1+49\)

\(A=\left(x-1\right)^2+49\ge49\)

Dấu "=" xảy ra khi:

\(x=1\)

\(B=12x-x^2\)

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

\(B=-x^2+12x-36+36\)

\(B=-\left(x^2-12x+36\right)+36\)

\(B=-\left(x-6\right)^2+36\le36\)

Dấu "=" xảy ra khi:

\(x=6\)

\(C=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)\)

\(C=\left[\left(x+1\right)\left(x-6\right)\right]\left[\left(x-2\right)\left(x-3\right)\right]\)

\(C=\left[x\left(x-6\right)+1\left(x-6\right)\right]\left[x\left(x-3\right)-2\left(x-3\right)\right]\)

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

\(C=\left(x^2-5x-6\right)\left(x^2-5x+6\right)\)

\(C=\left(x^2-5x\right)^2-36\ge-36\)

Dấu "=" xảy ra khi:

\(x^2-5x=0\)

\(\Rightarrow x\left(x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)