Tìm gtnn của A= 13x^2+y^2+4xy-2y-16x+2019 30/11/2021 Bởi Katherine Tìm gtnn của A= 13x^2+y^2+4xy-2y-16x+2019
Đáp án: `A=13x^2+y^2+4xy-2y-16x+2019` `=4x^2+9x^2+y^2+4xy-2y-4x-12x+2019` `=(4x^2+4xy+y^2-4x-2y+1)+(9x^2-12x+4)+2014` `=(2x+y-1)^2+(3x-2)^2+2014` Ta có: `(2x+y-1)^2>=0; (3x-2)^2>=0` `=> (2x+y-1)^2+(3x-2)^2>=0` `=> (2x+y-1)^2+(3x-2)^2+2014>=2014` Dấu “=” xảy ra `<=> ` $\left\{\begin{matrix}2x+y-1=0& \\3x-2=0& \end{matrix}\right.$ `=>` $\left\{\begin{matrix}x=\dfrac{2}{3}& \\y=-\dfrac{1}{3}& \end{matrix}\right.$ Vậy `A_(min)=2014 <=>` $\left\{\begin{matrix}x=\dfrac{2}{3}& \\y=-\dfrac{1}{3}& \end{matrix}\right.$ Bình luận
Đáp án: $\min A = 2014 \Leftrightarrow (x;y)=\left(\dfrac23;-\dfrac13\right)$ Giải thích các bước giải: $\begin{array}{l}\quad A = 13x^2+y^2+4xy-2y-16x+2019\\ \to A = (4x^2 + 4xy + y^2 – 4x – 2y + 1) + (9x^2 – 12x +4) + 2014\\ \to A = (2x +y -1)^2 + (3x – 2)^2 + 2014\\ \text{Ta có:}\\ \begin{cases}(2x+y-1)^2 \geq 0\quad \forall x;y\\(3x- 2)^2 \geq 0\quad \forall x\end{cases}\\ \text{Do đó:}\\ \quad (2x +y -1)^2 + (3x – 2)^2 + \geq 2014\\ \to A \geq 2014 \\ \text{Dấu = xảy ra}\,\,\Leftrightarrow \begin{cases}2x + y – 1 =0\\3x – 2 =0\end{cases} \Leftrightarrow \begin{cases}x = \dfrac23\\y = -\dfrac13\end{cases}\\ Vậy\,\,\min A = 2014 \Leftrightarrow (x;y)=\left(\dfrac23;-\dfrac13\right)\end{array}$ Bình luận
Đáp án:
`A=13x^2+y^2+4xy-2y-16x+2019`
`=4x^2+9x^2+y^2+4xy-2y-4x-12x+2019`
`=(4x^2+4xy+y^2-4x-2y+1)+(9x^2-12x+4)+2014`
`=(2x+y-1)^2+(3x-2)^2+2014`
Ta có: `(2x+y-1)^2>=0; (3x-2)^2>=0`
`=> (2x+y-1)^2+(3x-2)^2>=0`
`=> (2x+y-1)^2+(3x-2)^2+2014>=2014`
Dấu “=” xảy ra `<=> ` $\left\{\begin{matrix}2x+y-1=0& \\3x-2=0& \end{matrix}\right.$
`=>` $\left\{\begin{matrix}x=\dfrac{2}{3}& \\y=-\dfrac{1}{3}& \end{matrix}\right.$
Vậy `A_(min)=2014 <=>` $\left\{\begin{matrix}x=\dfrac{2}{3}& \\y=-\dfrac{1}{3}& \end{matrix}\right.$
Đáp án:
$\min A = 2014 \Leftrightarrow (x;y)=\left(\dfrac23;-\dfrac13\right)$
Giải thích các bước giải:
$\begin{array}{l}\quad A = 13x^2+y^2+4xy-2y-16x+2019\\ \to A = (4x^2 + 4xy + y^2 – 4x – 2y + 1) + (9x^2 – 12x +4) + 2014\\ \to A = (2x +y -1)^2 + (3x – 2)^2 + 2014\\ \text{Ta có:}\\ \begin{cases}(2x+y-1)^2 \geq 0\quad \forall x;y\\(3x- 2)^2 \geq 0\quad \forall x\end{cases}\\ \text{Do đó:}\\ \quad (2x +y -1)^2 + (3x – 2)^2 + \geq 2014\\ \to A \geq 2014 \\ \text{Dấu = xảy ra}\,\,\Leftrightarrow \begin{cases}2x + y – 1 =0\\3x – 2 =0\end{cases} \Leftrightarrow \begin{cases}x = \dfrac23\\y = -\dfrac13\end{cases}\\ Vậy\,\,\min A = 2014 \Leftrightarrow (x;y)=\left(\dfrac23;-\dfrac13\right)\end{array}$