Câu hỏi của Hoang Nguyen

Question

Tim gia tri nho nhat:

36 - 3x+$\dfrac{1}{2}$ $x^2$

Aki Tsuki · Accepted Answer

$A=36-3x+\dfrac{1}{2}x^2=\dfrac{1}{2}\left(x^2-6x+72\right)$

$=\dfrac{1}{2}\left[\left(x^2-6x+9\right)+63\right]=\dfrac{1}{2}\left[\left(x-3\right)^2+63\right]$

Có: $\left(x-3\right)^2\ge0\forall x\Rightarrow\left(x-3\right)^2+63\ge63$

$\dfrac{1}{2}\left[\left(x-3\right)^2+63\right]\ge\dfrac{1}{2}\cdot63=\dfrac{63}{2}$

Dấu ''='' xảy ra khi x = 3

Vậy $MIN_A=\dfrac{63}{2}\Leftrightarrow x=3$Câu trả lời: $A=36-3x+\dfrac{1}{2}x^2=\dfrac{1}{2}\left(x^2-6x+72\right)$

$=\dfrac{1}{2}\left[\left(x^2-6x+9\right)+63\right]=\dfrac{1}{2}\left[\left(x-3\right)^2+63\right]$

Có: $\left(x-3\right)^2\ge0\forall x\Rightarrow\left(x-3\right)^2+63\ge63$

$\dfrac{1}{2}\left[\left(x-3\right)^2+63\right]\ge\dfrac{1}{2}\cdot63=\dfrac{63}{2}$

Dấu ''='' xảy ra khi x = 3

Vậy $MIN_A=\dfrac{63}{2}\Leftrightarrow x=3$

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