Đang onl bằng điện thoại nên mình làm sơ sơ thôi nhé :((

A = ( x² - 3x + 9/4 ) + ( y² - 4y + 4 ) - 5/4

= ( x - 3/2 )² + ( y - 2 )² - 5/4 >= -5/4

Dấu = xảy ra <=> x = 3/2 ; y = 2

Vậy ...

B = ( x² - 2xy + y² ) + ( y² + 4y + 4 ) - 11

= ( x - y )² + ( y + 2 )² - 11 >= -11

Dấu = xảy ra <=> x = y = -2

Vậy ...

a) \(A=x^2+4y^2-3x-4y+5\)

\(=\left(x^2-3x+\frac{9}{4}\right)+\left(4y^2-4y+1\right)+\frac{7}{4}\)

\(=\left(x-\frac{3}{2}\right)^2+\left(2y-1\right)^2+\frac{7}{4}\)

Vì \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\); \(\left(2y-1\right)^2\ge0\forall y\)

\(\Rightarrow\left(x-\frac{3}{2}\right)^2+\left(2y-1\right)^2\ge0\forall x,y\)

\(\Rightarrow\left(x-\frac{3}{2}\right)^2+\left(2y-1\right)^2+\frac{7}{4}\ge\frac{7}{4}\forall x,y\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-\frac{3}{2}=0\\2y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\2y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\y=\frac{1}{2}\end{cases}}\)

Vậy \(minA=\frac{7}{4}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\y=\frac{1}{2}\end{cases}}\)

b) \(B=x^2+2y^2-2xy+4y-7\)

\(=\left(x^2-2xy+y^2\right)+\left(y^2+4y+4\right)-11\)

\(=\left(x-y\right)^2+\left(y+2\right)^2-11\)

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

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

\(\Rightarrow\left(x-y\right)^2+\left(y+2\right)^2-11\ge-11\forall x,y\)

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

Vậy \(minB=-11\)\(\Leftrightarrow x=y=-2\)

Ta có \(C=x^2+2y^2-2xy-4y+5=\left(x-2xy+y^2\right)+\left(y^2-4y+4\right)+1\)

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

Do \(\left(x-y\right)^2\ge0;\left(y-2\right)^2\ge0\Rightarrow C\ge1\)

Vậy GTNN của C là 1 khi \(\hept{\begin{cases}x-y=0\\y-2=0\end{cases}}\Leftrightarrow x=y=2\)

\(A=x^2+2y^2+2xy+2x-4y+2016\)

\(=x^2+y^2+y^2+2xy+2x+2y-6y+2016\)

\(=\left(x^2+2xy+y^2\right)+\left(y^2-6y+9\right)+\left(2x+2y\right)+2007\)

\(=\left(x+y\right)^2+\left(y-3\right)^2+2\left(x+y\right)+2007\)

\(=\left(x+y+1\right)^2+\left(y-3\right)^2+2006\)

Vì \(\hept{\begin{cases}\left(x+y+1\right)^2\ge0;\forall x,y\\\left(y-3\right)^2\ge0;\forall x,y\end{cases}}\)\(\Rightarrow\left(x+y+1\right)^2+\left(y-3\right)^2\ge0;\forall x,y\)

\(\Rightarrow\left(x+y+1\right)^2+\left(y-3\right)^2+2006\ge0+2006;\forall x,y\)

Hay \(A\ge2006;\forall x,y\)

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

Vậy \(A_{min}=2006\)\(\Leftrightarrow\hept{\begin{cases}x=2\\y=3\end{cases}}\)

Mình làm có gì sai hả @@ 

do em điểm cao qua mà

tích cho a đi

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

bàn làm giúp mình đk ko ạ!