\(A=\left|3-x\right|+8\ge8\)

\(minA=8\Leftrightarrow x=3\)

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

\(minB=-4\Leftrightarrow x=-2\)

\(A=\left|3-x\right|+8\ge8\forall x\)

Dấu '=' xảy ra khi x=3

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

Dấu '=' xảy ra khi x=-2

\(A = | x - 2019 | - | x - 2018 |\)

\(A = | x - 2019 | - | x - 2018 | \)\(\le\)\(| x - 2019 - x + 2018 |\)\(= | - 1 | = 1\)

\(Dấu " = " xảy ra\)\(\Leftrightarrow\)\(x - 2019 = 0 hoặc x - 2018 = 0\)

\(\Rightarrow\)\(x = 2019 hoặc x = 2018\)

\(Max A = 1 \)\(\Leftrightarrow\)\(x = 2019 hoặc x = 2018\)

Đặt \(A=\left|x-2018\right|+\left|x-2020\right|\)

\(\ge\left|\left(x-2018\right)+\left(2020-x\right)\right|=2\)

(Dấu "="\(\Leftrightarrow\left(x-2018\right)\left(2020-x\right)\ge0\)

\(\Leftrightarrow2018\le x\le2020\))

Vậy \(A_{min}=2\Leftrightarrow2018\le x\le2020\)

Đặt \(B=\left|x-2019\right|\ge0\)

(Dấu "="\(\Leftrightarrow x-2019=0\Leftrightarrow x=2019\))

Vậy \(B_{min}=0\Leftrightarrow x=2019\)

\(\Rightarrow\left|x-2018\right|+\left|x-2019\right|+\left|x-2020\right|\ge2\)

(Dấu "="\(\Leftrightarrow\hept{\begin{cases}2018\le x\le2020\\x=2019\end{cases}}\Leftrightarrow x=2019\))

Vậy \(BT_{min}=2\Leftrightarrow x=2019\)

\(A=\left|x-2019\right|-\left|x-2018\right|\)

Áp dụng BĐT \(\left|a\right|-\left|b\right|\le\left|a-b\right|\)ta có :

\(A\ge\left|x-2019-x+2018\right|=\left|-1\right|=1\)

Vậy ................

Nhầm Chỗ A 

Sửa thành \(A\le\left|x-2019-x+2018\right|=\left|-1\right|=1\)

\(|x-2019|+|x-2|\ge|x-2019+2-x|=2017\)

Dau "=" xay ra khi:

\(\left(x-2\right)\left(x-2019\right)\ge0\Leftrightarrow1\le x\le\frac{2019}{2}\)

tt

a) Đặt \(A=\frac{2018}{|x|+2019}\)

Vì \(|x|\ge0;\forall x\)

\(\Rightarrow|x|+2019\ge0+2019;\forall x\)

\(\Rightarrow\frac{2018}{|x|+2019}\le\frac{2018}{2019};\forall x\)

Hay \(A\le\frac{2018}{2019};\forall x\)

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

Vậy MIN \(A=\frac{2018}{2019}\Leftrightarrow x=0\)

b) Đặt \(B=\frac{|x|+2018}{-2019}\)

Vì \(|x|\ge0;\forall x\)

\(\Rightarrow|x|+2018\ge0+2018;\forall x\)

\(\Rightarrow\frac{|x|+2018}{-2019}\le\frac{-2018}{2019};\forall x\)

Hay \(B\le\frac{-2018}{2019};\forall x\)

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

Vạy MIN \(B=\frac{-2018}{2019}\Leftrightarrow X=0\)

a) Vì \(\sqrt{x-5}\) ≥0

⇒ \(\sqrt{x-5}+7\) ≥ 7

Min A=7⇔x-5=0

             ⇔x=5

b) Vì \(\sqrt{3x-5}\) ≥0

⇒ 8-\(\sqrt{3x-5}\) ≤8

Max=8⇔3x-5\(=\)0

           ⇔\(x=\dfrac{5}{3}\)