Question

Tìm a để M có giá trị nhỏ nhất: \(M=\dfrac{a^2-2a+2008}{a^2}\) với a khác 0.

Answer 1

\(M=\frac{a^2-2a+2008}{a^2}\)

\(M=\frac{a^2}{a^2}-\frac{2a}{a^2}+\frac{2008}{a^2}\)

\(M=1-\frac{2}{a}+\frac{2008}{a^2}\)

\(M=1-2\cdot\frac{1}{a}+2008\cdot\left(\frac{1}{a}\right)^2\)

Đặt \(\frac{1}{a}=x\)

Ta có :

\(M=1-2x+2008x^2\)

\(M=2008\left(x^2-x\cdot\frac{1}{1004}+\frac{1}{2008}\right)\)

\(M=2008\left(x^2-2\cdot x\cdot\frac{1}{2008}+\frac{1}{2008^2}+\frac{2007}{2008^2}\right)\)

\(M=2008\left[\left(x-\frac{1}{2008}\right)^2+\frac{2007}{2008^2}\right]\)

\(M=2018\left(x-\frac{1}{2008}\right)^2+\frac{2007}{2008}\ge\frac{2007}{2008}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=\frac{1}{2008}\)