Question

Tìm giá trị nhỏ nhất của x\(^2\)+3x+4

@Phùng Khánh Linh

Answer 1

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 \(x=\dfrac{-3}{2}\)

Answer 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

Answer 3

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

Do : \(\left(x+\dfrac{3}{2}\right)^2\) ≥ 0 ∀x

⇒ \(\left(x+\dfrac{3}{2}\right)^2\) + \(\dfrac{7}{4}\) ≥ \(\dfrac{7}{4}\)

⇒ B_Min = \(\dfrac{7}{4}\) ⇔ x = \(\dfrac{-3}{2}\)

Answer 4

\(x^2+3x+4\)

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

\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\)

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 : \(\left(x+\dfrac{3}{2}\right)^2=0=>x=-\dfrac{3}{2}\)

Answer 5

@ngonhuminh ; @Nhã Doanh

Answer 6

\(x^2+3x+4=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)

Vậy .............................

Answer 7

chú tuổi gì :))) đây là câu hỏi nhằm mục đích câu cmt à :)))))))