Đặt: A=3-10x²-4xy-4y²=3-(10x²+4xy+4y²)​=3-[9x²+(x²+4xy+4y²)]=3-[9x²+(x+2y)²]

Do [9x²+(x+2y)²]\(\ge\)0 với mọi x, y 

=> A=3-[9x²+(x+2y)²]\(\le\)3 với mọi x, y

=> GTLN của A là 3

Đạt được khi x=y=0

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

GTNN = 5

2) tuong tu 

GTLN la gi ?

Có A = 3 - 10x\(^2\) - 4xy - 4y\(^2\) 

         = 3 - ( 10x\(^2\) + 4xy + 4y\(^2\) )

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

         = 3 - \([\)( 3x )\(^2\) + ( x + 2y ) \(^2\) \(]\)

Đánh giá    ............

Dấu "=" xảy ra ..........

\(A=3-10x^2-4xy-4y^2\)

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

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

\(A=3-\left[\left(3x\right)^2+\left(x+2y\right)^2\right]\le3\forall x;y\)

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

Vậy \(A_{max}=3\Leftrightarrow x=y=0\)

Lời giải:

Ta có:
\(A=3-(10x^2+4xy+4y^2)=3-[9x^2+(x^2+4xy+4y^2)]\)

\(=3-[(3x)^2+(x+2y)^2]\)

Vì \((3x)^2\geq 0; (x+2y)^2\geq 0\Rightarrow (3x)^2+(x+2y)^2\geq 0, \forall x,y\)

\(\Rightarrow A=3-[(3x)^2+(x+2y)^2]\leq 3\)

Vậy $A_{\max}=3$. Dấu "=" xảy ra khi \((3x)^2=(2x+y)^2=0\Rightarrow x=y=0\)

a) \(5x^2-12xy+9y^2-4x+4=\left(4x^2-12xy+9y^2\right)+x^2-4x+4=\left(2x-3y\right)^2+\left(x-2\right)^2\ge0\)
b) \(-x^2-2y^2+12x-4y+7=-\left(x^2-12x+36\right)-2\left(y^2+2y+1\right)+45=-\left(x-6\right)^2-2\left(y+1\right)^2+45\le45\)

c)\(4y^2+10x^2+12xy+6x+7=\left(4y^2+12xy+9x^2\right)+x^2+6x+9-2=\left(2y+3x\right)^2+\left(x+3\right)^2-2\ge-2\)

d) \(3-10x^2-4xy-4y^2=3-\left(4y^2+4xy+x^2\right)-9x^2=-\left(2y+x\right)^2-9x^2+3\le3\)

e)\(x^2-5x+y^2-xy-4y+16=\left(\frac{1}{2}x^2-xy+\frac{1}{2}y^2\right)+\frac{1}{2}\left(x^2-10x+25\right)+\frac{1}{2}\left(y^2-8y+16\right)-\frac{9}{2}=\frac{1}{2}\left(x-y\right)^2+\frac{1}{2}\left(x-5\right)^2+\frac{1}{2}\left(y-4\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)Phần e) mới nghĩ đk v, tui biết đáp án sao do k xảy ra dấu bằng

\(B=9x^2-12x=\left(9x^2-12x+4\right)-4=\left(3x-2\right)^2-4\ge-4\)Vậy \(Min_B=-4\) khi \(3x-2=0\Rightarrow3x=2\Rightarrow x=\dfrac{2}{3}\)

\(D=3-10x^2-4xy-4y^2=3-\left(3x\right)^2-\left(x^2+4xy+4y^2\right)=3-\left(3x\right)^2-\left(x+2y\right)^2\le3\)Vậy \(Max_D=3\) khi \(\left[{}\begin{matrix}3x=0\\x+2y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\2y=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)