Đặt \(P=x^2+4y^2-4xy+2x-4y+9\)

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

\(P=\left(x-2y+1\right)^2+8\ge8\)

\(P_{min}=8\) khi \(x-2y+1=0\)

yiouoiyy

\(2x^2+2y^2+z^2+2xy+2xz+2yz+10x+6y+34=0\)

\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)

Vì \(\hept{\begin{cases}\left(x+y+z\right)^2\ge0\\\left(x+5\right)^2\ge0\\\left(y+3\right)^2\ge0\end{cases}}\)\(\Rightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+y+z\right)^2=0\\\left(x+5\right)^2=0\\\left(y+3\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y+z=0\\x+5=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x+y+z=0\\x=-5\\y=-3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-5\\y=-3\\z=8\end{cases}}}\)

\(A=2x^2+4y^2+4xy+2x+4y+9=\left(x^2+4y^2+4xy+2x+4y+1\right)+x^2+8\)

   \(=\left(x+2y+1\right)^2+x^2+8\ge8\)

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

Vậy \(Min\left(A\right)=8\Leftrightarrow\hept{\begin{cases}x=0\\y=-\frac{1}{2}\end{cases}}\)

d)  D = x⁴ - 6x² + 10

D = (X²)² - 2. x². 3 + 3² + 1

D = (x² - 3)² + 1

(x² - 3)²  >= 0 với mọi x

(x² - 3)² + 1 >=1 với moi5 x

Vậy GTNN của D là 1

Nãy lộn nhé,em làm lại:

\(D=\left(x^2+4xy+2x+4y^2+4y+1\right)+x^2+8\)

\(=\left[x^2+2x\left(2y+1\right)+\left(2y+1\right)^2\right]+x^2+8\)

\(=\left(x+2y+1\right)^2+x^2+8\ge8\)

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

Dạng này mình không quen cho lắm nên không chắc nha!

\(D=\left(x^2+4xy+2x+4y^2+4y+1\right)+8\)

\(=\left[x^2+2x\left(2y+1\right)+\left(2y+1\right)\right]+8\)

\(=\left(x+2y+1\right)^2+8\ge8\)

Dấu "=" xảy ra khi \(\left(x+2y+1\right)^2=0\Leftrightarrow2y+1=-x\)

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

\(=x^2-2x^2-x=-x^2-x=0\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

Thay vào D loại x = -1 suy ra x = 0 tức là y = -1/2

1) (x-1)² + (x- 4y)² + (y + 2)² +10 -1-4

GTNN = 5

2) tuong tu 

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

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

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

\(=2\left(x+\dfrac{2y+1}{2}\right)^2+2y^2+2y+\dfrac{17}{2}\)

\(=2\left(x+\dfrac{2y+1}{2}\right)^2+2\left(y+\dfrac{1}{2}\right)^2+8\ge8\)

Dấu '' = '' xảy ra khi: \(\Leftrightarrow\left\{{}\begin{matrix}y+\dfrac{1}{2}=0\\x+\dfrac{2y+1}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{2}\\x=0\end{matrix}\right.\)

Vậy: Min A = 8 khi \(x=0;y=-\dfrac{1}{2}\)