Question

Answer 1

Lời giải:

ĐKXĐ: $x\neq 1$
\(A=\frac{3(x^2-2x+1)-2(x-1)+1}{(x-1)^2}=\frac{3(x-1)^2-2(x-1)+1}{(x-1)^2}\)

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

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

Vậy $A_{\min}=2$ khi $x=2$

 

Answer 2

ĐKXĐ: \(x\ne1\)

Đặt \(x-1=t\ne0\Rightarrow x=t+1\)

\(A=\dfrac{3\left(t+1\right)^2-8\left(t+1\right)+6}{t^2}=\dfrac{3t^2-2t+1}{t^2}=\dfrac{1}{t^2}-\dfrac{2}{t}+3=\left(\dfrac{1}{t}-1\right)^2+2\ge2\)

\(A_{min}=2\) khi \(t=1\Rightarrow x=2\)