\(A=\left(x^2+2xy+y^2\right)+\left(y^2-6y+9\right)+2021\\ A=\left(x+y\right)^2+\left(y-3\right)^2+2021\ge2021\\ A_{min}=2021\Leftrightarrow\left\{{}\begin{matrix}x=-y=-3\\y=3\end{matrix}\right.\)

A=(x2+2xy+y2)+(y2−6y+9)+2021A=(x+y)2+(y−3)2+2021≥2021Amin=2021⇔{x=−y=−3y=3

a) \(A=x^2+2y^2+2xy+4x+6y+19\)

\(=\left[\left(x^2+2xy+y^2\right)+2.\left(x+y\right).2+4\right]+\left(y^2+2y+1\right)+14\)

\(=\left[\left(x+y\right)^2+2\left(x+y\right).2+2^2\right]+\left(y+1\right)^2+14\)

\(=\left(x+y+2\right)^2+\left(y+1\right)^2+14\ge14\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x+y+2=0\\y=-1\end{cases}}\Leftrightarrow x=y=-1\)

b)Đề có gì đó sai sai...

c) Tương tự câu b,em cũng thấy sai sai...HÓng cao nhân giải ạ!

b) \(P=2x^2+y^2+2xy-2y-4\)

\(\Leftrightarrow2P=4x^2+2y^2+4xy-4y-8\)

\(\Leftrightarrow2P=\left(4x^2+4xy+y^2\right)+\left(y^2-4y+4\right)-12\)

\(\Leftrightarrow2P=\left(2x+y\right)^2+\left(y-2\right)^2-12\ge-12\forall x;y\)

Có \(2P\ge-12\Leftrightarrow P\ge-6\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}2x+y=0\\y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\y=2\end{cases}}}\)

\(A=x^2-2xy+6y^2-12x+2y+45\)

\(A=\left(x^2-2xy+y^2-12x+12y+36\right)+\left(5y^2-10y+5\right)+4\)

\(A=\left[\left(x-y\right)^2-12.\left(x-y\right)+6^2\right]+5\left(y^2-2y+1\right)+4\)

\(A=\left(x-y-6\right)^2+5.\left(y-1\right)^2+4\)

Vì \(\left(x-y-6\right)^2\ge0\forall x,y\)

\(5.\left(y-1\right)^2\ge0\forall x,y\)

\(\Rightarrow A_{Min}=4\Leftrightarrow y=1,x=7\)

\(A=x^2+y^2+2xy+4x+4y+4+y^2+2y+1+14\)

\(A=\left(x+y+2\right)^2+\left(y+1\right)^2+14\ge14\)

\(\Rightarrow A_{min}=14\) khi \(\left\{{}\begin{matrix}y=-1\\x=-1\end{matrix}\right.\)

\(B=2\left(x^2+xy+\frac{y^2}{4}\right)+\frac{1}{2}\left(y^2-4y+4\right)-6\)

\(B=2\left(x+\frac{y}{2}\right)^2+\frac{1}{2}\left(y-2\right)^2-6\ge-6\)

\(\Rightarrow B_{min}=-6\) khi \(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

Câu c đề sai, sao vừa có 2xy lại có cả 4xy

\(2x^2+2y^2-2xy-6y+21\)

\(2A=4x^2+4y^2-4xy-12y+42\)

\(=4x^2-4xy+4y^2-12y+42\)

\(=4x^2-4xy+y^2+3y^2-12y+42\)

\(=\left(4x^2-4xy+y^2\right)+\left(3y^2-12y+42\right)\)

\(=\left(2x-y\right)^2+3\left(y^2-4x+4\right)+30\)

\(=\left(2x-y\right)^2+3\left(y-2\right)^2+30\ge30\)

Vậy GTNN là 30

Cho mk sủa lại tí :

\(2A=4x^2+4y^2-4xy-12y+42\)

\(=4x^2-4xy+4y^2-12+42\)

\(=4x^2-4xy+y^2+3y^2-12y+42\)

\(=\left(2x-y\right)^2+3\left(y-2\right)^2+30\ge30\)

\(\Rightarrow2A\ge30\Rightarrow A\ge15\Rightarrow\)GTNN là 15

*\(A=x^2+2y^2-2xy-4x-6y-3\)

\(A=x^2-2x\left(y+2\right)+\left(y^2+4y+4\right)+\left(y^2-10y+25\right)-32\)

\(A=x^2-2x\left(y+2\right)+\left(y+2\right)^2+\left(y-5\right)^2-32\)

\(A=\left(x-y-2\right)^2+\left(y-5\right)^2-32\ge-32\)

\(\Rightarrow Min_A=-32\Leftrightarrow x=7;y=5\)

* \(B=4x^2+2y^2-4xy+4x+6y+1\)

\(B=\left(2x\right)^2-\left(4xy+4x\right)+\left(y^2-2y+1\right)+\left(y^2+8y+16\right)-16\)\(B=\left(2x\right)^2-2.2x\left(y-1\right)+\left(y-1\right)^2+\left(y+4\right)^2-16\)\(B=\left(2x-y+1\right)^2+\left(y+4\right)^2-16\ge-16\)

\(\Rightarrow Min_B=-16\Leftrightarrow x=-\dfrac{5}{2};y=-4\)

Bạn chép thiếu đề à??

2x² + 2y² + 2xy - 6y + 21

= (x² + 2xy + y²) - 2(x + y) + 1 + (x² + 2x + 1) + (y² - 4y + 4) + 15

= (x + y)² - 2(x + y) + 1 + (x + 1)² + (y - 2)² + 15

= (x + y - 1)² + (x + 1)² + (y - 2)² + 15 \(\ge15\)

Vậy GTNN là 15 đạt được khi x = - 1, y = 2

phân tích đa thức có dạng m² + n ( n thuộc z)

bàn làm giúp mình đk ko ạ!