Question

[Nguyễn Quốc Huy]

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

Đặt t= x-1 => x= t+1

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

Đặt \(\frac{1}{t}\)= b \(\Rightarrow\)E= b²-2b+3= b²-2b+1+2= (b-1)²+2\(\ge\)2

Min E= 2 khi b-1= 0 \(\Rightarrow\)b= 1\(\Rightarrow\)\(\frac{1}{t}\)= 1\(\Rightarrow\)\(\frac{1}{x-1}\)= 1 \(\Leftrightarrow\)x-1= 1 \(\Rightarrow\)x=1