\(A=13x^2+y^2+4xy-2y-16x+2015\)

\(A=\left(4x^2-4x+1\right)+2y\left(2x-1\right)+y^2+\left(9x^2-12x+4\right)+2010\)

\(A=\left(2x-1\right)^2+2y\left(2x-1\right)+y^2+\left(3x-2\right)^2+2010\)

\(A=\left(2x-1+y\right)^2+\left(3x-2\right)^2+2010\)

Đến đây bạn tự làm nốt nhé~

không làm được thì ib

câu 2: gọi biểu thức là A đi

\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+ab=1.\left[\left(a+b\right)^2-3ab\right]+ab=\left(a+b\right)^2-2ab=1-2ab\)

\(\left(a-b\right)^2\ge0\Rightarrow a^2+b^2\ge2ab\Rightarrow\left(a+b\right)^2\ge4ab\Leftrightarrow ab\le\frac{1}{4}\)(chỗ 4ab là cộng 2 vế với 2ab đó)

\(\Leftrightarrow-ab\ge\frac{-1}{4}\Leftrightarrow-2ab\ge-\frac{1}{2}\Rightarrow1-2ab\ge\frac{1}{2}\Rightarrow A\ge\frac{1}{2}\Rightarrowđpcm\)

\(13x^2+y^2-4xy-16x+2y+2022\)

\(=\left(y-2x\right)^2+2\left(y-2x\right).1+1+9x^2-12x+4+2017\)

\(=\left(y-2x+1\right)^2+\left(3x-2\right)^2+2017\)

Vậy: Min là 2017 khi \(x=\dfrac{2}{3};y=\dfrac{1}{3}\)

a, \(2x^2-4xy+4y^2-6x\)

\(=x^2-2xy-2xy+4y^2+x^2-3x-3x+9-9\)

\(=\left(x-2y\right)^2+\left(x-3\right)^2-9\)

Với mọi giá trị của \(x;y\in R\) ta có:

\(\left(x-2y\right)^2+\left(x-3\right)^2-9\ge-9\)

Để \(\left(x-2y\right)^2+\left(x-3\right)^2-9=-9\) thì

\(\left\{{}\begin{matrix}\left(x-2y\right)^2=0\\\left(x-3\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3-2y=0\\x=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=1,5\\x=3\end{matrix}\right.\)

Vậy..............

b, \(z^2-4zt+5t^2-2t+13\)

\(=z^2-2zt-2zt+4t^2+t^2-t-t+1+12\)

\(=\left(z-2t\right)^2+\left(t-1\right)^2+12\)

Với mọi giá trị của \(z;t\in R\) ta có:

\(\left(z-2t\right)^2+\left(t-1\right)^2+12\ge12\)

Để \(\left(z-2t\right)^2+\left(t-1\right)^2+12=12\) thì

\(\left\{{}\begin{matrix}\left(z-2t\right)^2=0\\\left(t-1\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}z-2=0\\t=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}z=2\\t=1\end{matrix}\right.\)

Vậy...............

Câu c tường tự !!!

a,Đặt A= \(2x^2-4xy+4y^2-6x\)

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

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

\(=2\left[x^2-2x\left(y+\dfrac{3}{2}\right)+\left(y+\dfrac{3}{2}\right)^2\right]+4y^2-\left(y+\dfrac{3}{2}\right)^2\)

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

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

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

Với mọi giá trị của x;y ta có:

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

\(\Rightarrow2\left(x-y-\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2-3\ge-3\)

Vậy Min A = -3 khi \(\left\{{}\begin{matrix}x-y-\dfrac{3}{2}=0\\y-\dfrac{1}{2}=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}-\dfrac{3}{2}=0\\y=\dfrac{1}{2}\end{matrix}\right.\)

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

b, Đặt B = \(z^2-4zt+5t^2-2t+13\)

\(=\left(z^2-4zt+4t^2\right)+\left(t^2-2t+1\right)+12\)

\(=\left(z-2t\right)^2+\left(t-1\right)^2+12\)

Với mọi giá trị của z;t ta có:

\(\left(z-2t\right)^2\ge0;\left(t-1\right)^2\ge0\)

\(\Rightarrow\left(z-2t\right)^2+\left(t-1\right)^2+12\ge12\)

Vậy Min B = 12 khi \(\left\{{}\begin{matrix}z-2t=0\\t-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}z-2=0\\t=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}z=2\\t=1\end{matrix}\right.\)

c, Đặt C = \(16x^2-8x+y^2-2y\)

\(=\left(16x^2-8x+1\right)+\left(y^2-2y+1\right)-2\)

\(=\left(4x-1\right)^2+\left(y-1\right)^2-2\)

Với mọi giá trị x;y ta có:

\(\left(4x-1\right)^2\ge0;\left(y-1\right)^2\ge0\)

\(\Rightarrow\left(4x-1\right)^2+\left(y-1\right)^2-2\ge-2\)

Vậy Min C = -2 khi \(\left\{{}\begin{matrix}4x-1=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x=1\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=1\end{matrix}\right.\)