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

GTNN = 5

2) tuong tu 

\(C=2x^2+5y^2+4xy+8x-4y-100 \)

\(C=\left(x^2+8x+16\right)+\left(y^2-4y+4\right)+\left(x^2+4xy+4y^2\right)-120\)

\(C=\left(x+4\right)^2+\left(y-2\right)^2+\left(x+2y\right)^2-120\ge-120\)

Vậy GTNN của C là -120 khi x = -4; y = 2

\(C=x^2+4xy+4y^2+x^2+8x+16+y^2-4y+4-120\)

\(=\left(x+2y\right)^2+\left(x+4\right)^2+\left(y-2\right)^2-120\ge-120\)

vậy GTNN của C là -120 khi \(x=-4;y=2\)

\(P=\left(x^2-4xy+4y^2\right)+2\left(x-2y\right)+x^2-4x+2019\)

\(=\left(x-2y\right)^2+2\left(x-2y\right)+1+\left(x-2\right)^2+2014\)

\(=\left(x-2y+1\right)^2+\left(x-2\right)^2+2014\ge2014\)

Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x=2\\x=2y-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=\frac{3}{2}\end{matrix}\right.\)

Vậy...

\(P=2x^2+4y^2-4xy-2x-4y+2019\)

\(P=\left(x^2-4xy+4y^2\right)+2\left(x-2y\right)+1+\left(x-2\right)^2+2014\) ( Bước này mình làm hơi tắt , cái này bạn chỉ cần chú ý để tách ra thôi )

\(P=\left(x-2y+1\right)^2+\left(x-2\right)^2+2014\ge2014\)

Dấu '' = '' xảy ra

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}3-2y=0\\x=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3}{2}\\x=2\end{matrix}\right.\)

Vậy Min \(P=2014\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3}{2}\\x=2\end{matrix}\right.\)

Bạn tên gì

:))

\(A=2x^2+4y^2+4xy+10x+12y+18\)

\(A=x^2+4xy+4y^2+6x+12y+9+x^2+4x+4+5\)

\(A=\left(x+2y^2\right)+2.3\left(x+2y\right)+9+\left(x+2\right)^2+5\)

\(A=\left(x+2y+3\right)^2+\left(x+2\right)^2+5\)

Do \(\hept{\begin{cases}\left(x+2y+3\right)^2\ge0\forall x\\\left(x+2\right)^2\ge0\forall x\end{cases}}\)

\(\Leftrightarrow\left(x+2y+3\right)^2+\left(x+2\right)^2+5\ge5\)

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

\(\Rightarrow A_{min}=5\Leftrightarrow\hept{\begin{cases}x=-2\\y=-\frac{1}{2}\end{cases}}\)

Chúc bạn học tốt !!!

a) \(M=10x^2+6y+4y^2+4xy+2\)

\(=\left(10x^2+4xy+\dfrac{2}{5}y^2\right)+\left(\dfrac{18}{5}y^2+6y+\dfrac{5}{2}\right)-\dfrac{1}{2}\)

\(=10\left(x^2+\dfrac{2}{5}xy+\dfrac{1}{25}y^2\right)+\dfrac{18}{5}\left(y^2+\dfrac{5}{3}y+\dfrac{25}{36}\right)-\dfrac{1}{2}\)

\(=10\left(x+\dfrac{1}{5}y\right)^2+\dfrac{18}{5}\left(y+\dfrac{5}{6}\right)^2-\dfrac{1}{2}\ge-\dfrac{1}{2}\)

Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{5}y=0\\y+\dfrac{5}{6}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{5}{6}\end{matrix}\right.\)

b) \(H=-x^2+2xy-4y^2+2x+10y-8\)

\(=-x^2+2x\left(y+1\right)-\left(y^2+2y+1\right)-\left(3y^2-12y+7\right)\)

\(=-x^2+2x\left(y+1\right)-\left(y+1\right)^2-3\left(y^2-4y+4\right)+5\)

\(=-\left(x-y-1\right)^2-3\left(y-2\right)^2+5\le5\)

Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-y-1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)

c) \(K=2x^2+2xy-2x+2xy+y^2\)

bn xem lại cái đề nhé, sao lại có 2 lần 2xy

Câu c đúng đề mà

=x² +4y²+4xy+1+2x+4y+x²+9

=(x+2y)²+2(x+2y)+1+x²+9

=(x+2y+1)²+x²+9

có (x+2y+1)²≥0 với mọi x,y

x²≥0 với mọi x

⇒(x+2y+1)²+x² ≥0với mọi x,y

⇒(x+2y+1)²+x²+9≥9với mọi x,y

⇒

ta có :

A = 2x²+4y²+4xy+2x+4y+9 = 2x²+2x+4y²+4y+4xy+9

= 2x(x+1)+4y(y+1)+4xy+9

= 2x(x+1)+4y(y+x+1)+9

= (x+1)(2x+4y²)+9

=> A lớn hơn hoặc bằng 9

=> min A là 9

Ta có: A = 2x² + 4y² - 4xy - 4x - 4y + 15

= (x² - 4xy + 4y²) + 2(x - 2y) + 1 + (x² - 6x + 9) + 5

= (x - 2y)² + 2(x - 2y) + 1 + (x - 3)² + 5

= (x - 2y + 1)² + (x - 3)² + 5 \(\ge\)5 \(\forall\)x; y

Daaus "=" xảy ra <=> \(\left\{{}\begin{matrix}x-2y+1=0\\x-3=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}y=\frac{x+1}{2}\\x=3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}y=2\\x=3\end{matrix}\right.\)

Vậy MinA = 5 khi x = 3 và y = 2

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

\(=\left(y-x+1\right)^2+2\left(y-2\right)^2+5\ge5\)

Vậy MIN=5 khi \(\left\{{}\begin{matrix}y=2\\x=3\end{matrix}\right.\)