Question

tim gia tri nho nhat cua bieu thuc A=2022.Ix^2+1I+2023

Answer 1

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 \(x=0\)

#\(Toru\)

Answer 2

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

mà \(x^2+1\ge1\forall x\)

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