Câu hỏi của Phúc Nguyễn

Question

Tìm giá trị nhỏ nhất của biểu thức sau:

A=2x^2-6x-2xy+y^2+10

Tìm giá trị lớn nhất của biểu thức sau:

A=5+2xy+14y-x^2-5y^2-2x

Khánh Ngọc · Accepted Answer

1. $A=2x^2-6x-2xy+y^2+10$

$\Leftrightarrow A=\left(x^2-2xy+y^2\right)+\left(x^2-6x+9\right)+1$

$\Leftrightarrow A=\left(x-y\right)^2+\left(x-3\right)^2+1$

Vì $\left(x-y\right)^2\ge0$ ; $\left(x-3\right)^2\ge0$$\forall x;y$

$\Rightarrow A=\left(x-y\right)^2+\left(x-3\right)^2+1\ge1$

Dấu "=" xảy ra $\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\left(x-3\right)^2=0\end{matrix}\right.\Leftrightarrow x=y=3$

Vậy minA = 1 $\Leftrightarrow x=y=3$

2. $A=5+2xy+14y-x^2-5y^2-2x$

$\Leftrightarrow A=-\left(x^2-2xy+y^2+2x-2y+1\right)-\left(4y^2-12y+9\right)+15$

$\Leftrightarrow A=-\left(x-y+1\right)^2-\left(2y-3\right)^2+15$

Vì $\left\{{}\begin{matrix}\left(x-y+1\right)^2\ge0\\left(2y-3\right)^2\ge0\end{matrix}\right.$$\forall x;y$

$\Rightarrow A=-\left(x-y+1\right)^2-\left(2y-3\right)^2+15\le15$

Dấu "=" xảy ra $\Leftrightarrow\left\{{}\begin{matrix}\left(x-y+1\right)^2=0\\left(2y-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\y=\frac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\y=\frac{3}{2}\end{matrix}\right.$

Vậy maxA = 15 $\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\y=\frac{3}{2}\end{matrix}\right.$Câu trả lời: 1. $A=2x^2-6x-2xy+y^2+10$

$\Leftrightarrow A=\left(x^2-2xy+y^2\right)+\left(x^2-6x+9\right)+1$

$\Leftrightarrow A=\left(x-y\right)^2+\left(x-3\right)^2+1$

Vì $\left(x-y\right)^2\ge0$ ; $\left(x-3\right)^2\ge0$$\forall x;y$

$\Rightarrow A=\left(x-y\right)^2+\left(x-3\right)^2+1\ge1$

Dấu "=" xảy ra $\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\left(x-3\right)^2=0\end{matrix}\right.\Leftrightarrow x=y=3$

Vậy minA = 1 $\Leftrightarrow x=y=3$

2. $A=5+2xy+14y-x^2-5y^2-2x$

$\Leftrightarrow A=-\left(x^2-2xy+y^2+2x-2y+1\right)-\left(4y^2-12y+9\right)+15$

$\Leftrightarrow A=-\left(x-y+1\right)^2-\left(2y-3\right)^2+15$

Vì $\left\{{}\begin{matrix}\left(x-y+1\right)^2\ge0\\left(2y-3\right)^2\ge0\end{matrix}\right.$$\forall x;y$

$\Rightarrow A=-\left(x-y+1\right)^2-\left(2y-3\right)^2+15\le15$

Dấu "=" xảy ra $\Leftrightarrow\left\{{}\begin{matrix}\left(x-y+1\right)^2=0\\left(2y-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\y=\frac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\y=\frac{3}{2}\end{matrix}\right.$

Vậy maxA = 15 $\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\y=\frac{3}{2}\end{matrix}\right.$

Nguyễn Gia Hưng · Answer

1. A=2x2−6x−2xy+y2+10A=2x2−6x−2xy+y2+10

⇔A=(x2−2xy+y2)+(x2−6x+9)+1⇔A=(x2−2xy+y2)+(x2−6x+9)+1

⇔A=(x−y)2+(x−3)2+1⇔A=(x−y)2+(x−3)2+1

Vì (x−y)2≥0(x−y)2≥0 ; (x−3)2≥0(x−3)2≥0∀x;y∀x;y

⇒A=(x−y)2+(x−3)2+1≥1⇒A=(x−y)2+(x−3)2+1≥1

Dấu "=" xảy ra ⇔{(x−y)2=0(x−3)2=0⇔x=y=3⇔{(x−y)2=0(x−3)2=0⇔x=y=3

Vậy minA = 1 ⇔x=y=3⇔x=y=3

2. A=5+2xy+14y−x2−5y2−2xA=5+2xy+14y−x2−5y2−2x

⇔A=−(x2−2xy+y2+2x−2y+1)−(4y2−12y+9)+15⇔A=−(x2−2xy+y2+2x−2y+1)−(4y2−12y+9)+15

⇔A=−(x−y+1)2−(2y−3)2+15⇔A=−(x−y+1)2−(2y−3)2+15

Vì {(x−y+1)2≥0(2y−3)2≥0{(x−y+1)2≥0(2y−3)2≥0∀x;y∀x;y

⇒A=−(x−y+1)2−(2y−3)2+15≤15⇒A=−(x−y+1)2−(2y−3)2+15≤15

Dấu "=" xảy ra ⇔{(x−y+1)2=0(2y−3)2=0⇔{x−y=−1y=32⇔{x=12y=32⇔{(x−y+1)2=0(2y−3)2=0⇔{x−y=−1y=32⇔{x=12y=32

Vậy maxA = 15 ⇔{x=12y=32Câu trả lời: 1. A=2x2−6x−2xy+y2+10A=2x2−6x−2xy+y2+10

⇔A=(x2−2xy+y2)+(x2−6x+9)+1⇔A=(x2−2xy+y2)+(x2−6x+9)+1

⇔A=(x−y)2+(x−3)2+1⇔A=(x−y)2+(x−3)2+1

Vì (x−y)2≥0(x−y)2≥0 ; (x−3)2≥0(x−3)2≥0∀x;y∀x;y

⇒A=(x−y)2+(x−3)2+1≥1⇒A=(x−y)2+(x−3)2+1≥1

Dấu "=" xảy ra ⇔{(x−y)2=0(x−3)2=0⇔x=y=3⇔{(x−y)2=0(x−3)2=0⇔x=y=3

Vậy minA = 1 ⇔x=y=3⇔x=y=3

2. A=5+2xy+14y−x2−5y2−2xA=5+2xy+14y−x2−5y2−2x

⇔A=−(x2−2xy+y2+2x−2y+1)−(4y2−12y+9)+15⇔A=−(x2−2xy+y2+2x−2y+1)−(4y2−12y+9)+15

⇔A=−(x−y+1)2−(2y−3)2+15⇔A=−(x−y+1)2−(2y−3)2+15

Vì {(x−y+1)2≥0(2y−3)2≥0{(x−y+1)2≥0(2y−3)2≥0∀x;y∀x;y

⇒A=−(x−y+1)2−(2y−3)2+15≤15⇒A=−(x−y+1)2−(2y−3)2+15≤15

Dấu "=" xảy ra ⇔{(x−y+1)2=0(2y−3)2=0⇔{x−y=−1y=32⇔{x=12y=32⇔{(x−y+1)2=0(2y−3)2=0⇔{x−y=−1y=32⇔{x=12y=32

Vậy maxA = 15 ⇔{x=12y=32

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