3: 

Ta có: \(\left(2x+1\right)^2\ge0\forall x\)

\(\Leftrightarrow\left(2x+1\right)^2+2021\ge2021\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)

a) Ta có : \(A=\left|x+1\right|+\left|y-2\right|\)

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

\(=\left|x+y-1\right|=\left|5-1\right|=\left|4\right|=4\)

Dấu "=" xảy ra <=> (x + 1)(y - 2) \(\ge\)0

Vậy Min A = 4 <=>  (x + 1)(y - 2) \(\ge\)0

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

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

\(\Rightarrow A\ge2\)

Dấu '' = '' xảy ra khi

\(\left(2x-1\right)\left(3-2x\right)\ge0\)

\(\Leftrightarrow\frac{1}{2}\le x\le\frac{3}{2}\)

Vậy Min A = 2 \(\Leftrightarrow\frac{1}{2}\le x\le\frac{3}{2}\)

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

\(=\left|2x-1\right|^2-3\left|2x-1\right|+2\)

\(=\left|2x+1\right|^2-2\left|2x-1\right|.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\frac{1}{4}\)

\(\rightarrow B=\left(\left|2x-1\right|-\frac{3}{2}\right)^2-\frac{1}{4}\ge0-\frac{1}{4}=-\frac{1}{4}\)

\(\Rightarrow Min=\frac{-1}{4}\)

Dấu " =" xảy ra \(\Leftrightarrow\left|2x-1\right|=\frac{3}{2}\Rightarrow\left[{}\begin{matrix}x=\frac{5}{4}\\x=-\frac{1}{4}\end{matrix}\right.\)

a, \(A=\left|x+1\right|+\left|y-2\right|\)

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

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

Dấu " = " sảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\3-x\ge0\end{matrix}\right.\Leftrightarrow-1\le x\le3\)

A = /2*-5-3/1+/2*-5