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 

a) B= \(\frac{x^2-x+1}{x^2+2x+1}\)=\(\frac{x^2-x+1}{\left(x+1\right)^2}\)

Đặt t = x+1 => x= t-1 => B= \(\frac{\left(t-1\right)^2-\left(t-1\right)+1}{t^2}\)=\(\frac{t^2-2t+1-t+1+1}{t^2}\)

B= \(\frac{t^2-3t+3}{t^2}\)=\(\frac{t^2}{t^2}\)\(-\frac{3t}{t^2}\)\(+\frac{3}{t^2}\)

Đặt a = \(\frac{1}{t}\)=> B= 3a²-3a+1= 3(a²-a+\(\frac{1}{3}\))= 3(a²-2a.\(\frac{1}{2}\)+\(\frac{1}{4}\)+\(\frac{1}{12}\))= 3(a-\(\frac{1}{2}\))²+\(\frac{1}{4}\)\(\ge\)\(\frac{1}{4}\)

Min của E= \(\frac{1}{4}\)khi a-\(\frac{1}{2}\)= 0 => a= \(\frac{1}{2}\)=> \(\frac{1}{t}\)=\(\frac{1}{2}\)\(\Leftrightarrow\)\(\frac{1}{x+1}\)=\(\frac{1}{2}\)\(\Leftrightarrow\)x+1=2 => x= 1