Tìm GTNN của biểu thức A = (x²+ 2x + 2023)/x² , (x khác 0)

\(A=\frac{x^2+2x+2023}{x^2}\)

\(A=\frac{2023x^2+2x.2023+2023^2}{2023x^2}\)

\(A=\frac{2022x^2+x^2+2x.2023+2023^2}{2023x^2}\)

\(A=\frac{2022}{2023}+\frac{\left(x+2023\right)^2}{2023x^2}\)

Ta có: \(\frac{\left(x+2023\right)^2}{2023x^2}\ge0\ ∀x\ne0\)

<=> \(\frac{2022}{2023}+\frac{\left(x+2023\right)^2}{2023x^2}\ge\frac{2022}{2023}\ ∀x\ne0\)

<=> \(A\ge\frac{2022}{2023}\ ∀x\ne0\)

Dấu "=" xảy ra <=> x + 2023 = 0 <=> x = - 2023

Vậy \(A_{\min }=\frac{2022}{2023}\) <=> x = - 2023

\(A=\frac{x^2+2x+2023}{x^2}\)

\(A=\ 1+\frac{2}{x}+\frac{2023}{x^2}\)

Đặt \(t=\frac{1}{x}\) (x khác 0)

<=> A = 2023t² + 2t +1

\(A=2023\left(t^2+\frac{2}{2023}t+\frac{1}{2023}\right)\)

\(A=2023\left(t^2+\frac{2}{2023}t+\frac{1}{2023^2}-\frac{1}{2023^2}+\frac{1}{2023}\right)\)

\(A=2023\left(t^2+\frac{2}{2023}t+\frac{1}{2023^2}\right)-\frac{1}{2023}+1\)

\(A=2023\left(t+\frac{1}{2023}\right)^2+\frac{2022}{2023}\)

Ta có: \(2023\left(t+\frac{1}{2023}\right)^2\ge0\ ∀x\)

<=> \(A=2023\left(t+\frac{1}{2023}\right)^2+\frac{2022}{2023}\ge\frac{2022}{2023}\ ∀x\)

<=> \(A\ \ge\frac{2022}{2023}\ ∀x\)

Vậy \(A_{\min}=\frac{2022}{2023}\) <=> \(t=-\frac{1}{2023}\) hay x = - 2023

Tham khảo chuyên đề tìm GTNN và GTLN tại https://vndoc.com/chuyen-de-tim-gtln-gtnn-cua-bieu-thuc-196098