Tìm giá trị nhỏ nhất của x\(^2\)+3x+4
Đặt: \(A=x^2+3x+4=x^2+2.x.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{7}{4}\)
\(A=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\)
Ta có: \(\left(x+\dfrac{3}{2}\right)\ge0\)
\(\Rightarrow A=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)
\(\Rightarrow Min_A=\dfrac{7}{4}\) khi \(x=\dfrac{-3}{2}\)
\(A=x^2+3x+4=\left(x^2+2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}\right)+\dfrac{7}{4}=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\)
Vì: \(\left(x+\dfrac{3}{2}\right)^2\ge0\forall x\Rightarrow\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)
Dấu ''='' xảy ra khi x = -3/2
Vậy Min_A = 7/4 khi x = -3/2
\(B=x^2+3x+4=x^2+2.\dfrac{3}{2}x+\dfrac{9}{4}+4-\dfrac{9}{4}=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\)
Do : \(\left(x+\dfrac{3}{2}\right)^2\) ≥ 0 ∀x
⇒ \(\left(x+\dfrac{3}{2}\right)^2\) + \(\dfrac{7}{4}\) ≥ \(\dfrac{7}{4}\)
⇒ BMin = \(\dfrac{7}{4}\) ⇔ x = \(\dfrac{-3}{2}\)
\(x^2+3x+4\)
\(=\left(x^2+2x.\dfrac{3}{2}+\dfrac{9}{4}\right)+\dfrac{7}{4}\)
\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\)
Ta có : \(\left(x+\dfrac{3}{2}\right)^2\ge0\forall x\)
=> \(\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\forall x\)
Dấu "=" xảy ra khi : \(\left(x+\dfrac{3}{2}\right)^2=0=>x=-\dfrac{3}{2}\)
\(x^2+3x+4=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)
Vậy .............................
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