Question

[Đỗ Thị Phương Anh]

Tim gtri nho nhat cua

A= x^2 - 5x + 12

B= 2x^2 -14x + 5

Answer 1

[Trần Quốc Lộc]

\(A=x^2-5x+12\\ A=x^2-5x+\dfrac{25}{4}+\dfrac{23}{4}\\ A=\left(x^2-5x+\dfrac{25}{4}\right)+\dfrac{23}{4}\\ A=\left[x^2-2\cdot x\cdot\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]+\dfrac{23}{4}\\ A=\left(x-\dfrac{5}{2}\right)^2+\dfrac{23}{4}\\ Do\text{ }\left(x-\dfrac{5}{2}\right)^2\ge0\forall x\\ \Rightarrow A=\left(x-\dfrac{5}{2}\right)^2+\dfrac{23}{4}\ge\dfrac{23}{4}\forall x\\ \text{Dấu "=" xảy ra khi : }\\ \left(x-\dfrac{5}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{5}{2}=0\\ \Leftrightarrow x=\dfrac{5}{2}\\ \text{Vậy }A_{\left(Min\right)}=\dfrac{23}{4}\text{ }khi\text{ }x=\dfrac{5}{2}\)

\(B=2x^2-14x+5\\ \\ A=2x^2-14x+\dfrac{49}{2}-\dfrac{39}{2}\\ A=\left(2x^2-14x+\dfrac{49}{2}\right)-\dfrac{39}{2}\\ A=2\left(x^2-7x+\dfrac{49}{4}\right)-\dfrac{39}{2}\\ A=\left[x^2-2\cdot x\cdot\dfrac{7}{2}+\left(\dfrac{7}{2}\right)^2\right]-\dfrac{39}{2}\\ A=\left(x-\dfrac{7}{2}\right)^2-\dfrac{39}{2}\\ Do\text{ }\left(x-\dfrac{7}{2}\right)^2\ge0\forall x\\ \Rightarrow A=\left(x-\dfrac{7}{2}\right)^2-\dfrac{39}{2}\ge-\dfrac{39}{2}\forall x\\ \text{Dấu "=" xảy ra khi : }\\ \left(x-\dfrac{7}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{7}{2}=0\\ \Leftrightarrow x=\dfrac{7}{2}\\ \text{Vậy }B_{\left(Min\right)}=-\dfrac{39}{2}\text{ }khi\text{ }x=\dfrac{7}{2}\)

Answer 2

[Trần Quốc Lộc]

\(B=2x^2-14x+5\\ B=2x^2-14x+\dfrac{49}{2}-\dfrac{39}{2}\\ B=\left(2x^2-14x+\dfrac{49}{2}\right)-\dfrac{39}{2}\\ B=2\left(x^2-7x+\dfrac{49}{4}\right)-\dfrac{39}{2}\\ B=2\left[x^2-2\cdot x\cdot\dfrac{7}{2}+\left(\dfrac{7}{2}\right)^2\right]-\dfrac{39}{2}\\ B=2\left(x-\dfrac{7}{2}\right)^2-\dfrac{39}{2}\\ \)

Do \(\left(x-\dfrac{7}{2}\right)^2\ge0\forall x\)

\(\Rightarrow2\left(x-\dfrac{7}{2}\right)^2\ge0\forall x\)

\(\Rightarrow B=2\left(x-\dfrac{7}{2}\right)^2-\dfrac{39}{2}\ge-\dfrac{39}{2}\forall x\)

Dấu \("="\) xảy ra khi :

\(\left(x-\dfrac{7}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{7}{2}=0\\ \Leftrightarrow x=\dfrac{7}{2}\)

Vậy \(B_{\left(Min\right)}=-\dfrac{39}{2}\) khi \(x=\dfrac{7}{2}\)

Do máy bị lỗi nên câu B bị trục trặc.

Mk xin lỗi.