bạn phải chơi doll thì mới làm dược

A = x² - 2xy + 3y² - 2x + 1997

= ( x² - 2xy + y² - 2x + 2y + 1 ) + ( 2y² - 2y + 1/2 ) + 3991/2

= [ ( x² - 2xy + y² ) - ( 2x - 2y ) + 1 ] + 2( y² - y + 1/4 ) + 3991/2

= [ ( x - y )² - 2( x - y ) + 1² ] + 2( y - 1/2 )² + 3991/2

= ( x - y - 1 )² + 2( y - 1/2 )² + 3991/2 ≥ 3991/2 ∀ x, y

Dấu "=" xảy ra <=> x = 3/2 ; y = 1/2

=> MinA = 3991/2 <=> x = 3/2 ; y = 1/2

\(4A=12x^2+12y^2+4z^2+20xy-12yz-12zx-8x-8y+12\)

\(=9x^2+9y^2+4z^2+18xy-12yz-12zx+2\left(x^2+y^2+4-4x-4y+2xy\right)+x^2+y^2-2xy+4\)

\(=\left(3x+3y-2z\right)^2+2\left(x+y-2\right)^2+\left(x-y\right)^2+4\ge4\)

Dấu \(=\)khi \(\hept{\begin{cases}3x+3y-2z=0\\x+y-2=0\\x-y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y=1\\z=3\end{cases}}\).

Vậy \(minA=1\)khi \(x=y=1,z=3\).

\(A=3x^2+3y^2+z^2+5xy-3yz-3xz-2x-2y+3\)

\(=\left(z-\frac{3}{2}x-\frac{3}{2}y\right)^2+\frac{3}{4}\left(x^2y^2+\frac{2}{3}xy-\frac{8}{3}x-\frac{8}{3}y\right)+3\)

\(=\left(z-\frac{3}{2}x-\frac{3}{2}y\right)^2+\frac{3}{4}[\left(x+\frac{y}{3}-\frac{4}{3}\right)^2+\frac{8}{9}y^2-\frac{16}{9}y-\frac{16}{9}]\)

\(=\left(z-\frac{3}{2}x-\frac{3}{2}y\right)^2+\frac{3}{y}[\left(x+\frac{y}{3}-\frac{4}{3}\right)^2+\frac{8}{9}\left(y-1\right)^2-\frac{2y}{9}]+3\)

\(=\left(z-\frac{3}{2}x-\frac{3}{2}y\right)^2+\frac{3}{y}[\left(x+\frac{y}{3}-\frac{4}{3}\right)^2+\frac{8}{9}\left(y-1\right)^2]+1\)

\(\Leftrightarrow A\ge1\Leftrightarrow MinA=1\)

Dấu '' = '' xảy ra khi:

\(\hept{\begin{cases}z-\frac{3}{2}x-\frac{3}{2}y=0\\y-1=0\\x+\frac{y}{3}-\frac{4}{3}=0\end{cases}}\)

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

3 nha banj

xin hỏi bạn có viết lộn không, vế trái không có Z mà tại sao vế phải lại xuất hiện Z vậy