Question

tìm min của biểu thức A=\(\frac{x^2-x+2}{x^2}\) với x\(\ne0\)

Answer 1

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

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

\(A=2\left[\left(\frac{1}{x}\right)^2-2\cdot\frac{1}{x}\cdot\frac{1}{4}+\left(\frac{1}{4}\right)^2+\frac{7}{16}\right]\)

\(A=2\left(\frac{1}{x}-\frac{1}{4}\right)^2+2\cdot\frac{7}{16}\)

\(A=2\left(\frac{1}{x}-\frac{1}{4}\right)^2+\frac{7}{8}\) \(\ge\frac{7}{8}\forall x\)

\(A=\frac{7}{8}\) \(\Leftrightarrow\frac{1}{x}=\frac{1}{4}\Leftrightarrow x=4\)

Vậy Min A \(=\frac{7}{8}\) \(\Leftrightarrow x=4\)