Question

[Hải Anh Bùi]

tính giá trị nhỏ nhất 2x^2-6x+8

Answer 1

[Phương' ss ngốc]

A= 2x^2 - 6x 
=2.(x^2 - 2. 3/2 .x + 9/4) - 9/2 
= 2.( x- 3/2) ^2 - 9/2 >= -9/2 
--> Min A = -9/2 khi x =3/2

Answer 2

[tth_new]

Đặt \(A=2x^2-6x+8\)

\(=2x^2-6x+\frac{9}{2}+\frac{7}{2}\)

\(=2\left(x^2-3x+\frac{9}{4}\right)+\frac{7}{2}\)

\(=2\left(x-\frac{3}{2}\right)^2+\frac{7}{2}\ge\frac{7}{2}\) (do \(2\left(x-\frac{3}{2}\right)^2\ge0\forall x\))

Dấu "=" xảy ra \(\Leftrightarrow2\left(x-\frac{3}{2}\right)^2=0\Leftrightarrow x=\frac{3}{2}\)

Vậy \(A_{min}=\frac{7}{2}\Leftrightarrow x=\frac{3}{2}\)