Question

\(y=\frac{x^2+7}{\left(x-1\right)^2}\) Tìm giá trị nhỏ nhất của y

Answer 1

\(y=\frac{\left(x^2-2x+1\right)+2.\left(x-1\right)+8}{\left(x-1\right)^2}=\frac{\left(x-1\right)^2}{\left(x-1\right)^2}+\frac{2\left(x-1\right)}{\left(x-1\right)^2}+\frac{8}{\left(x-1\right)^2}=1+\frac{2}{x-1}+\frac{8}{\left(x-1\right)^2}\)

Đặt t = \(\frac{1}{x-1}\)

=> y = 1 + 2t + 8t² = 8.(t² + 2.\(\frac{1}{8}\). t + \(\left(\frac{1}{8}\right)^2\) ) - \(\frac{1}{8}\) + 1 = 8. (t + \(\frac{1}{8}\))² + \(\frac{7}{8}\) \(\ge\) 8.0 + \(\frac{7}{8}\) = \(\frac{7}{8}\) với mọi t

=> Min y = \(\frac{7}{8}\) khi t + \(\frac{1}{8}\) = 0 <=> t = -\(\frac{1}{8}\)<=>\(\frac{1}{x-1}\)  = \(\frac{1}{8}\) <=> x - 1 = 8 <=> x = 9