Ta có :  

\(A=3x^2+2y^2+2xy-10x-10y+2030\)

\(A=3x^2+2\left(y-5\right)x+2y^2-10y+2030\)

\(\Leftrightarrow3x^2+2\left(y-5\right)x+2y^2-10y+2030+A\ge0\)

\(\Delta'=\left(y-5\right)^2-3\left(2y^2-10y+2030-A\right)\ge0\)

\(\Leftrightarrow-5y^2+20y-6065+3A\ge0\)

\(\Leftrightarrow3A\ge5y^2-20y+6065=5\left(y^2-4y+4\right)+6045\)

\(\Leftrightarrow3A\ge5\left(y-2\right)^2+6045\)

\(\Leftrightarrow A\ge\frac{5}{3}\left(y-2\right)^2+2015\ge2015\)

Vậy  \(MinA=2015\Leftrightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}\)

\(A=3x^2+2\left(y-5\right)x+2y^2-10y+2030\)

\(\Leftrightarrow3x^2+2\left(y-5\right)x+2y^2-10y+2030-A=0\)

Để tồn tại x, y thỏa mãn, ta phải có:

\(\Delta'=\left(y-5\right)^2-3\left(2y^2-10y+2030-A\right)\ge0\)

\(\Leftrightarrow-5y^2+20y-6065+3A\ge0\)

\(\Leftrightarrow3A\ge5y^2-20y+6065=5\left(y^2-4y+4\right)+6045\)

\(\Leftrightarrow3A\ge5\left(y-2\right)^2+6045\Rightarrow A\ge\dfrac{5}{3}\left(y-2\right)^2+2015\ge2015\)

\(\Rightarrow A_{min}=2015\) khi \(y=2\Rightarrow x=1\)

Làm theo kiểu lớp 8 thì như sau:

\(A=2y^2+2y\left(x-5\right)+3x^2-10x+2030\)

\(A=2\left(y^2+2y.\dfrac{\left(x-5\right)}{2}+\left(\dfrac{x-5}{2}\right)^2\right)+\dfrac{5}{2}\left(x^2-2x+1\right)+2025\)

\(A=2\left(y+\dfrac{x-5}{2}\right)^2+\dfrac{5}{2}\left(x-1\right)^2+2025\ge2025\)

\(\Rightarrow A_{min}=2025\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}x-1=0\\y+\dfrac{x-5}{2}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

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

bàn làm giúp mình đk ko ạ!

3x² + 2y² + 2xy - 10x - 10y + 15 = 0

<=> 6x² + 4y² + 4xy - 20x - 20y + 30 = 0

<=> (4x² + 4xy + y²) - 10(2x + y) + 25 + (5y² - 10xy + 5) = 0

<=> (2x + y)² - 10(2x + y) + 25 + 5(y - 1)² = 0

<=> (2x + y - 5)² + 5(y - 1)² = 0

<=> \(\hept{\begin{cases}2x+y-5=0\\y-1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=\frac{5-y}{2}\\y=1\end{cases}}\)

<=> \(\hept{\begin{cases}x=2\\y=1\end{cases}}\)

\(3x^2+2y^2+2xy-10x-10y+15=0\)

\(\Rightarrow\left(x^2+2xy+y^2-6x-6y+9\right)+\left(2x^2-4x+2\right)+\left(y^2-4y+4\right)=0\)

\(\Rightarrow\left(x+y-3\right)^2+2\left(x-1\right)^2+\left(y-2\right)^2=0\)

mà \(\left(x+y-3\right)^2\ge0\forall x,y\)

\(2\left(x-1\right)^2\ge0\forall x\)

\(\left(y-2\right)^2\ge0\forall y\)

\(\Rightarrow\hept{\begin{cases}x+y-3=0\\x-1=0\\y-2=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}}\)

Yêu cầu?

2030

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

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

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