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...
Đọc tiếp
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
