Question

tim MIN cua bieu thuc sau

\(D=\frac{x^2-3x+3}{x^2-2x+1}\)

Answer 1

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

Đặt: x-1=y=>x=y+1. Ta có:

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

Đặt: \(\frac{1}{y}=t\Rightarrow D=1-t+t^2\ge\frac{3}{4}\\ D=\frac{3}{4}\Leftrightarrow\left(t-\frac{1}{2}\right)^2=0\Rightarrow t=\frac{1}{2}\)

\(t=\frac{1}{2}\Leftrightarrow\frac{1}{y}=\frac{1}{2}\Rightarrow y=2\Leftrightarrow x-1=2\Rightarrow x=3\)

Vậy minD=\(\frac{3}{4}\Leftrightarrow x=3\)

Answer 2

D=\(\frac{x.x-3x+3}{x.x-2x+1}\)

D=\(\frac{x.\left(x-3\right)+3}{x.\left(x-2\right)+1}\)

D=\(\frac{x-3+3}{x-2+2}\)(Chia cả tử và mẫu cho x lần)

D=\(\frac{x}{x}\)

D=1