`|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 = \(\dfrac{1}{2}\)

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

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

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

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 \(\Leftrightarrow x=0\)

Đặt `B = |x - 1| + |x - 2| + |x - 3| + |x - 4|`

`= (|x - 1| + |x - 4|) + (|x - 2| + |x - 3|)`

`= (|x - 1| + |4 - x|) + (|x - 2| + |3 - x|)`

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

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

\(\Rightarrow A\ge4+15=19\)

hay MinA = 19

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}\left(x-1\right)\left(4-x\right)\ge0\\\left(x-2\right)\left(3-x\right)\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(x-4\right)\le0\\\left(x-2\right)\left(x-3\right)\le0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}1\le x\le4\\2\le x\le3\end{matrix}\right.\Rightarrow2\le x\le3\)

Vậy MinA = 19 tại \(2\le x\le3\).

a ) Để \(\dfrac{3}{-x^2+2x+4}\) đạt GTlN thì :

\(-x^2+2x+4\) phải đạt GTNN ( chắc ai cũng biết )

Ta có :

\(-x^2+2x+4\)

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

\(=-\left(x-1\right)^2-5\)

Tới đây chắc bạn hỉu rồi nhỉ ?

a) \(A=\left(6x-1\right)^2+2017\)

Vì \(\left(6x-1\right)^2\ge0\)

Nên \(\left(6x-1\right)^2+2017\ge2017\)

Vậy GTNN của A=2017 khi \(6x-1=0\Leftrightarrow x=\dfrac{1}{6}\)

c) \(C=15+\left|2x-1\right|\)

Vì \(\left|2x-1\right|\ge0\)

Nên \(\left|2x-1\right|+15\ge15\)

Vậy GTNN của C=15 khi \(2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)

d) \(D=\left(x-1\right)^2+\left(2x-y\right)^2+3\)

Vì \(\left(x-1\right)^2+\left(2x-y\right)^2\ge0\)

Nên \(\left(x-1\right)^2+\left(2x-y\right)^2+3\ge3\)

Vậy GTNN của D=3 khi \(\left\{{}\begin{matrix}x-1=0\\2x-y=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\2.1-y=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)