\(F=2x^2-10x+20=2\left(x-\frac{5}{2}\right)^2+\frac{15}{2}\ge\frac{15}{2},\forall x\)
\(\Rightarrow minF=\frac{15}{2}\Leftrightarrow x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)
F = 2x2 - 10x + 20
= 2( x2 - 5x + 25/4 ) + 15/2
= 2( x - 5/2 )2 + 15/2 ≥ 15/2 ∀ x
Đẳng thức xảy ra <=> x - 5/2 = 0 => x = 5/2
=> MinF = 15/2 <=> x = 5/2
Bài giải
\(F=2x^2-10x+20=2\left(x^2-5x+\frac{25}{4}\right)+\frac{15}{2}=2\left(x-\frac{5}{2}\right)^2+\frac{15}{2}\ge\frac{15}{2}\forall x\)
Dấu " = " xảy ra khi : \(2\left(x-\frac{5}{2}\right)^2=0\text{ }\Rightarrow\text{ }x=\frac{5}{2}\)
Vậy \(Min_F=\frac{15}{2}\text{ khi }x=\frac{5}{2}\)
