Question

tìm giá trị nhỏ nhất : A=x^2-4x+1

Answer 1

\(A=x^2-4x+1=\left(x-2\right)^2-3\ge-3\)

\(\Rightarrow Min_A=-3\Leftrightarrow x=2\)

Answer 2

Giải:

\(A=x^2-4x+1\)

\(\Leftrightarrow A=x^2-4x+4-3\)

\(\Leftrightarrow A=\left(x^2-4x+4\right)-3\)

\(\Leftrightarrow A=\left(x-2\right)^2-3\ge-3;\forall x\)

\(\Leftrightarrow A_{Min}=-3\)

\("="\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy ...

Answer 3

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

Ta có:\(\left(x-2\right)^2\ge0\\ \Rightarrow\left(x-2\right)^2-3\ge-3\\\)

Vậy GTNN A=-3 ⇔x=2