Ta có : /2x-1/ >=0

Gtnn cuả /2x-1/ = 0 đạt tại x = 1/2

===> Gtnn của C là 0 + 2.1/2 + 6 = 7

Bài 1:

\(A=\left(x^2+2x+1\right)+1=\left(x+1\right)^2+1\ge1\)

\(A_{min}=1\) khi \(x+1=0\Leftrightarrow x=-1\)

\(B=\left(x-3\right)^2\ge0\)

\(B_{min}=0\) khi \(x=3\)

\(C=2\left(x^2-2.\frac{3}{2}x+\frac{9}{4}\right)+\frac{9}{2}=2\left(x-\frac{3}{2}\right)^2+\frac{9}{2}\ge\frac{9}{2}\)

\(C_{min}=\frac{9}{2}\) khi \(x=\frac{3}{2}\)

\(D=\left(x^2-2.\frac{1}{2}x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+\frac{3}{4}\)

\(D=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

\(D_{min}=\frac{3}{4}\) khi \(\left\{{}\begin{matrix}x=\frac{1}{2}\\y=-3\end{matrix}\right.\)

Bài 2:

\(A=-\left(x^2-4x+4\right)-1=-\left(x-2\right)^2-1\le-1\)

\(A_{max}=-1\) khi \(x=2\)

\(B=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

\(B_{max}=7\) khi \(x=2\)

\(C=-\left(x^2-2.\frac{1}{2}x+\frac{1}{4}\right)+\frac{1}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

\(C_{max}=\frac{1}{4}\) khi \(x=\frac{1}{2}\)

\(D=-\left(x^2-2x+1\right)-\left(y^2-4y+4\right)+11\)

\(D=-\left(x-1\right)^2-\left(y-2\right)^2+11\le11\)

\(D_{max}=11\) khi \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

\(E=-\frac{1}{2}\left(4x^2-4x+1\right)-\frac{9}{2}=-\frac{1}{2}\left(2x-1\right)^2-\frac{9}{2}\le-\frac{9}{2}\)

\(E_{max}=-\frac{9}{2}\) khi \(x=\frac{1}{2}\)

\(2x+\left|2x-5\right|=2x+\left|5-2x\right|\ge2x+5-2x=5.\Rightarrow A_{min}=5.\text{Dâu "=" xay }ra\Leftrightarrow2x-5\ge0\Leftrightarrow x\le2,5\)

\(M=\left|x\right|+\left|x-1\right|=\left|x\right|+\left|1-x\right|\ge x+1-x=1\Rightarrow M_{min}=1.\text{Dâu "=" xay ra}\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\1-x\ge0\end{matrix}\right.\Leftrightarrow0\le x\le1\)

\(A=x-\sqrt{x}\Leftrightarrow A+\frac{1}{4}=x-\sqrt{x}+\frac{1}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\Rightarrow A+\frac{1}{4}\ge0\Rightarrow A_{min}=\frac{-1}{4}.\text{Dâus "=" xay ra khi:}x=\frac{1}{4}\)

Bài 1:

Sửa đề :v

\(B=x\left(x-3\right)\left(x-1\right)\left(x-4\right)\)

\(B=\left(x^2-4x\right)\left(x^2-4x+3\right)\)

Đặt \(x^2-4x=t\)

\(B=t\left(t+3\right)\)

\(B=t^2+3t=t^2+2\cdot t\cdot\frac{3}{2}+\frac{9}{4}-\frac{9}{4}=\left(t+\frac{3}{2}\right)^2-\frac{9}{4}\ge\frac{-9}{4}\forall t\)

Dấu "=" xảy ra \(\Leftrightarrow t=\frac{-3}{2}\Leftrightarrow x^2-4x=\frac{-3}{2}\Leftrightarrow x=\frac{4\pm\sqrt{10}}{2}\)

Bài 2: Mình nghĩ nên sửa đề tìm min \(A=\left|2x\right|+\left|2x-5\right|\)

Bài 3:

\(M=\left|x\right|+\left|x-1\right|\)

\(M=\left|x\right|+\left|1-x\right|\ge\left|x+1-x\right|=1\)

Dấu "=" xảy ra \(\Leftrightarrow x\left(1-x\right)\ge0\Leftrightarrow0\le x\le1\)

Bài 4:

\(A=x-\sqrt{x}\)

Do điều kiện \(x\ge0\)

\(\Rightarrow A\ge0+0=0\)

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

F = 2( x²+ 6x/2 +9/4) +3 -9/2

GTNN F = -3/2

`|2x-1|>=0`

`=>4|2x-1|>=0`

`=>4|2x-1|+3>=3`

Dâu "=" `<=>X=1/2`

\(A=4\left|2x-1\right|+3\ge0+3=3\)

\(\Rightarrow A_{min}=3\Leftrightarrow2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)

\(4\left|2x-1\right|\ge0\left(\forall x\right)=>4\left|2x-1\right|+3\ge3\)

dấu= xảy ra <=>2x-1=0<=>x\(=\dfrac{1}{2}\)

\(=>A\ge3\)

vậy min A=3

Vì 4|2x - 1| \(\ge0\forall x\in R\Rightarrow A=4\left|2x-1\right|+3\ge3\forall x\in R\)

Dấu "=" xảy ra khi : \(4\left|2x-1\right|=0\Leftrightarrow x=\dfrac{1}{2}\)

Vậy MinA = 3 khi x = \(\dfrac{1}{2}\)

\(A=4\cdot\left|2x-1\right|+3\ge0+3=3\)

\(A_{min}=3\)

\(\Leftrightarrow2x-1=0\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

a ) \(A=\left|x+1\right|+\left|x+2\right|-2x+3\ge2x+3-2x+3=6\)

Dấu " = " xảy ra khi \(\left(x+2\right)\left(x+1\right)\ge0\)

b ) 

\(B=\left|2x+3\right|+\left|1-2x\right|\ge\left|2x+3+1-2x\right|=4\)

Dấu " = " xảy ra khi \(\left(2x+3\right)\left(1-2x\right)\ge0\)

c )

\(C=\left|x-1\right|+\left|x-2\right|+\left|x-2\right|\ge\left|x-1\right|+\left|2-x\right|\ge\left|x-1+2-x\right|=1\)

Dấu " = " xảy ra khi \(x=2\)