a) Từ M = x − 3 2 2 + 31 4 ≥ 31 4 ⇒ M min = 31 4 ⇔ x = 3 2 .  

b) Ta có N = ( x   +   2 y ) 2   +   ( y   –   2 ) 2   +   ( x   +   4 ) 2   –   120   ≥   -   120 .

Tìm được N min  = -120 Û x = -4 và y = 2.

Ta có : A = (2 - x)(x + 4)

= 2x - x² + 8 - 4x

= -x² - 6x + 8 

= -(x² + 6x) + 8

= -(x² + 6x + 9 - 9) + 8

= -(x² + 6x + 9) + 9 + 8

A = -(x + 3)² + 17

Vì - (x + 3)² \(\le0\forall x\)

Nên : A = -(x + 3)² + 17 \(\le17\forall x\)

Vậy A_max = 17 khi x = -3

a ) \(C=5x-3x^2+2\)

\(=-3\left(x^2-\dfrac{5}{3}x-\dfrac{2}{3}\right)\)

\(=-3\left(x^2-2x.\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{49}{36}\right)\)

\(=-3\left[\left(x-\dfrac{5}{6}\right)^2-\dfrac{49}{36}\right]\)

\(=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{49}{12}\le\dfrac{49}{12}\forall x\)

Dấu " = " xảy ra \(\Leftrightarrow x-\dfrac{5}{6}=0\Leftrightarrow x=\dfrac{5}{6}\)

Vậy GTLN của C là : \(\dfrac{49}{12}\Leftrightarrow x=\dfrac{5}{6}\)

b ) \(D=-8x^2+4xy-y^2+3\)

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

\(=-\left(2x-y\right)^2-4x^2+3\le3\forall x;y\)

Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}2x-y=0\\4x^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=y\\x^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=y\\x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=0\end{matrix}\right.\)

Vậy GTLN của D là : \(3\Leftrightarrow x=y=0\)

\(A=x^2-8x+16+x^2+4xy+4y^2+y^2+4y+4+2004\)

\(=\left(x-4\right)^2+\left(x+2y\right)^2+\left(y+2\right)^2+2004\ge2004\)

Dấu ''='' xảy ra khi x = 4 ; y = -2 

Ta có:

\(C=2x^2+3y^2+4xy-8x-2y+18\)

\(C=2\left(x^2+2xy+y^2\right)+y^2-8x-2y+18\)

\(C=2[\left(x+y\right)^2-4\left(x+y\right)+4]+\left(y^2+6y+9\right)+1\)

\(C=2\left(x+y-2\right)^2+\left(y+3\right)^2+1\ge1\)

Dấu "=" xảy ra \(\Leftrightarrow x+y=2\)và \(y=-3\)

Hay x = 5 , y = -3

-8x²+4xy-y²+10=10-(4x²-4xy+y²)-4x²=10-(2x-y)²-(2x)²

vi-(2x-y)²-(2x)² ≤0

=>10-(2x-y)²-(2x)²≤10

dau bang say ra khi (2x-y)²-(2x)²=0 

vậy gái trị nhỏ nhất là:10

\(Q=-8x^2+4xy-y^2+10\)<=>\(Q=10-4x^2+4xy-y^2-4x^2\)

<=>\(Q=10-\left[\left(2x^2\right)-4xy+y^2\right]-\left(2x\right)^2\)<=>\(Q=10-\left(2x-y\right)^2-\left(2x\right)^2\)

<=>\(Q=10-\left[\left(2x-y\right)^2+\left(2x\right)^2\right]\)

Vì \(\hept{\begin{cases}\left(2x-y\right)^2\ge0\\\left(2x\right)^2\ge0\end{cases}\Leftrightarrow\left(2x-y\right)^2+\left(2x\right)^2\ge0}\)\(\Leftrightarrow-\left[\left(2x-y\right)^2+\left(2x\right)^2\right]\le0\)

\(\Leftrightarrow Q=10-\left[\left(2x-y\right)^2+\left(2x\right)^2\right]\le10\)

=>Q_max=10 <=> \(\left(2x-y\right)^2=\left(2x\right)^2=0\)<=>\(2x-y=2x=0\) <=>\(x=y=0\)

Vậy Q_max=10 tại x=y=0

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

có \(\dfrac{1}{xy}=\dfrac{2}{2xy}\left(2\right)\)

từ(1)(2)=>A=\(\dfrac{1}{1-2xy}+\dfrac{2}{2xy}\ge\dfrac{\left(1+\sqrt{2}\right)^2}{1}=\left(1+\sqrt{2}\right)^2\)

=>Min A=(1+\(\sqrt{2}\))^2

b, ta có : \(x+y=1=>2x+2y=2\)

\(B=\dfrac{1}{x^2+y^2}+\dfrac{3}{4xy}=\dfrac{4}{4x^2+4y^2}+\dfrac{6}{8xy}\)\(\ge\dfrac{\left(2+\sqrt{6}\right)^2}{\left(2x+2y\right)^2}\)

\(=\dfrac{\left(2+\sqrt{6}\right)^2}{2^2}=\dfrac{5+2\sqrt{6}}{2}\)=>\(B\ge\dfrac{5+2\sqrt{6}}{2}\)

=>\(MinB=\dfrac{5+2\sqrt{6}}{2}\)