Question

tìm GTNN \(A=\frac{3x^2-8x+6}{x^2-2x+1}\)

Answer 1

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Answer 2

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Answer 3

\(A=\frac{3x^2-8x+6}{x^2-2x+1}\)

\(A=\frac{3x^2-8x+6}{\left(x-1\right)^2}\)

Đặt \(x-1=y\left(y\ne0\right)\)

\(\Rightarrow x=y+1\)

\(A=\frac{3\left(y+1\right)^2-8\left(y+1\right)+6}{y^2}\)

\(A=\frac{3\left(y^2+2y+1\right)-8y-8+6}{y^2}\)

\(A=\frac{3y^2+6y+3-8y-8+6}{y^2}\)

\(A=\frac{3y^2-2y+1}{y^2}\)

\(A=\frac{3y^2}{y^2}-\frac{2y}{y^2}+\frac{1}{y^2}\)

\(A=3-\frac{2}{y}+\frac{1}{y^2}\)

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

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

Mà \(\left(\frac{1}{y}-1\right)^2\ge0\)

\(\Rightarrow A\ge2\)

Dấu "=" xảy ra khi :

\(\frac{1}{y}-1=0\)

\(\Leftrightarrow\frac{1}{y}=1\)

\(\Leftrightarrow y=1\)

Mà : \(x=y+1\Rightarrow x=2\)

Vậy \(A_{Min}=2\Leftrightarrow x=2\)