Ta có: \(x^2\ge0\forall x\)

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

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

\(\Rightarrow2022\cdot\left|x^2+1\right|\ge2022\forall x\)

\(\Rightarrow2022\cdot\left|x^2+1\right|+2023\ge4045\forall x\)

Dấu \("="\) xảy ra \(\Leftrightarrow2022\cdot\left|x^2+1\right|=2022\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=1\\x^2+1=-1\end{matrix}\right.\)

\(\Leftrightarrow x^2+1=1\) (do \(x^2+1>0\forall x\) )

\(\Leftrightarrow x=0\)

\(Vậy:\)\(Min_A=4045\) khi \(x=0\)

#\(Toru\)

\(A=2022.\left|x^2+1\right|+2023\\ A=2022.\left(x^2+1\right)+2023\)

mà \(x^2+1\ge1\forall x\)

\(\Rightarrow GTNN\left(A\right)=2022+2023=4045\) \(\forall x\in R\)

Bài 2: 

a: \(C=-\left|2x-1,5\right|< =0\)

Dấu '=' xảy ra khi x=0,75

b: \(D=-\left|3x+6\right|+5\le5\)

Dấu '=' xảy ra khi x=-2

ta có:

\(\left|x-1\right|+\left|x-2\right|+\left|y-3\right|+\left|x-4\right|\)

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

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

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

\(\ge3\)

Dấu "=" xả ra khi \(\hept{\begin{cases}\left(x-1\right)\left(4-x\right)\ge0\\\left|x-2\right|=0\\\left|y-3\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}1\le x\le4\cdot\\x=2\left(TM\cdot\right)\\y=3\end{cases}}\)

Vậy \(x=2;y=3\)

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

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

X x 4 - 10 = 3

X x 4 = 3 + 10

X x 4 = 13

x = 13 : 4

x = \(\frac{13}{4}\)

Ta có:\(\left|x+1\right|\ge0;\left|x-2\right|\ge0;\left|x+7\right|\ge0\)

\(\Rightarrow\left|x+1\right|+\left|x-2\right|+\left|x+7\right|\ge0\)

\(\Rightarrow5x-10\ge0\)

\(\Rightarrow5x\ge10\)

\(\Rightarrow x\ge2\)

\(\Rightarrow\left|x+1\right|=x+1\)

  \(\left|x-2\right|=x-2\)

  \(\left|x+7\right|=x+7\)

Ta có:\(\left|x+1\right|+\left|x-2\right|+\left|x+7\right|=5x-10\)

\(\Rightarrow x+1+x-2+x+7=5x-10\)

\(\Rightarrow\)\(3x+6=5x-10\)

\(\Rightarrow6+10=5x-3x\)

\(\Rightarrow2x=16\)

\(\Rightarrow x=8\)

Vậy x=8 thỏa mãn