Question

[Toàn Nguyễn Đức]

Tìm giá trị nhỏ nhất của biểu thức N = \(2x^2+y^2+2xy-4x-2y+3\)

Answer 1

[_Guiltykamikk_]

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

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

\(N=\left[\left(x+y\right)^2-2\left(x+y\right)+1\right]+\left(x^2-2x+1\right)+1\)

\(N=\left(x+y-1\right)^2+\left(x-1\right)^2+1\)

Mà  \(\left(x+y-1\right)\ge0\forall x;y\)

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

\(\Rightarrow N\ge1\)

Dấu "=" xảy ra khi :

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

Vậy  \(N_{Min}=1\Leftrightarrow\left(x;y\right)=\left(1;0\right)\)

Answer 2

[JiYoonMin]

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

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

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

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

\(Do\)\(\left(x+y+1\right)^2\)\(\ge\)\(0\)\(\forall\)\(x\)\(;\)\(y\)

\(x^2\)\(\ge\)\(0\)\(\forall\)\(x\)

=.>\(\left(x+y+1\right)^2+x^2+2\)\(\ge\)\(2\)\(\forall\)\(x\)\(;\)\(y\)

=>\(N\)\(\ge\)\(2\)\(\forall\)\(x\)\(;\)\(y\)

Dấu = xảy ra khi: 

\(\hept{\begin{cases}\left(x+y+1\right)^2=0\\x^2=0\end{cases}}\)

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

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

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

Vậy \(N_{min}\)\(=\)\(2\)khi \(y=-1\)\(;\)\(x=0\)

Chúc pạn họk tốt~~~!!! :3