Question

Tìm giá trị nhỏ nhất của biểu thức: 2x² - 2x + 3

Answer 1

\(2x^2-2x+3\)

\(=\left(x^2-2x+1\right)+x^2+2\)

\(=\left(x-1\right)^2+x^2+2\)

Vì \(\left(x-1\right)^2\ge0\) \(_{\forall x}\)

\(x^2\ge0\) \(\forall x\)

\(\Rightarrow\left(x-1\right)^2+x^2\ge0\) \(\forall x\)

\(\Rightarrow\left(x-1\right)^2+x^2+2\ge2\) \(\forall x\)

Dấu '' = '' xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy Min \(2x^2-2x+3=2\Leftrightarrow x=1.\)

Answer 2

A= 2x² - 2x +3

=\(2x^2-2x+\dfrac{1}{2}+\dfrac{5}{2}\)

=\(\left(2x^2-2x+\dfrac{1}{2}\right)+\dfrac{5}{2}\)

=2(x²-x+\(\dfrac{1}{4}\) )+\(\dfrac{5}{2}\)

=\(2\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{2}\)

=>A≥\(\dfrac{5}{2}\)

=>MinA=\(\dfrac{5}{2}\) khi

x-\(\dfrac{1}{2}\)=0

=>x=\(\dfrac{1}{2}\)

vậy minA=\(\dfrac{5}{2}\) khi x=\(\dfrac{1}{2}\)

Answer 3

A= 2x² - 2x + 3

A = 2( x² - x ) + 3

A = 2( x² -2.\(\dfrac{1}{2}x+\dfrac{1}{4}\)) + 3 - \(\dfrac{1}{2}\)

A = 2( x - \(\dfrac{1}{2}\))² + \(\dfrac{5}{2}\)

Do : 2( x - \(\dfrac{1}{2}\))² ≥ 0 ∀x

=>2( x - \(\dfrac{1}{2}\))² + \(\dfrac{5}{2}\) ≥ \(\dfrac{5}{2}\)∀x

=> A_min = \(\dfrac{5}{2}\) ⇔ x = \(\dfrac{1}{2}\)