Bài 2: 

a: \(=-\left(x^2+2x-100\right)\)

\(=-\left(x^2+2x+1-101\right)\)

\(=-\left(x+1\right)^2+101< =101\)

Dấu = xảy ra khi x=-1

b: \(=-3\left(x^2-\dfrac{1}{3}x\right)\)

\(=-3\left(x^2-2\cdot x\cdot\dfrac{1}{6}+\dfrac{1}{36}-\dfrac{1}{36}\right)\)

\(=-3\left(x-\dfrac{1}{6}\right)^2+\dfrac{1}{12}< =\dfrac{1}{12}\)

Dấu = xảy ra khi x=1/6

c: \(=-\left(3x^2+4y^2-18x+8y-12\right)\)

\(=-\left(3x^2-18x+27+4y^2+8y+4-43\right)\)

\(=-3\left(x-3\right)^2-4\left(y+1\right)^2+43< =43\)

Dấu = xảy ra khi x=3 và y=-1

a: \(\Leftrightarrow x^2-2x+1+y^2+4y+4=0\)

=>(x-1)^2+(y+2)^2=0

=>x=1 và y=-2

b: \(\Leftrightarrow2x^2+2y^2-16x+32+16y+32=0\)

\(\Leftrightarrow2\left(y-4\right)^2+2\left(x+4\right)^2=0\)

=>y=4; x=-4

\(A=-x^2+6x-10=-\left(x^2-6x+9\right)-1=-\left(x-3\right)^2-1\le-1\)

Vậy GTLN của A là -1 khi x = 3

\(B=-2x^2-4x-10=-2\left(x^2+2x+1\right)-8=-2\left(x+1\right)^2-8\le-8\)

Vậy GTLN của B là -8 khi x = -1

\(C=-2x^2+3x-10=-2\left(x^2-\frac{3}{2}x+\frac{9}{16}\right)-\frac{71}{8}=-2\left(x-\frac{3}{4}\right)^2-\frac{71}{8}\le-\frac{71}{8}\)

Vậy GTLN của C là \(-\frac{71}{8}\)khi x = \(\frac{3}{4}\)

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

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

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

Vậy GTLN của D là -5 khi x = 1; y = -2

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

\(a,A=-x^2+6x-10\)

\(=-x^2+6x-9-1\)

\(=-\left(x^2-6x+9\right)-1\)

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

Ta có: \(-\left(x-3\right)^2\le0\forall x\)

\(\Rightarrow-\left(x-3\right)^2-1\le-1\forall x\)

=> Max A =-1 tại \(-\left(x-3\right)^2=0\Rightarrow x=3\)

cn lại lm tg tự 

=.= hok tốt!!

\(-\left(2x^2+y^2+2xy-4x-2y-5\right)\\ \\ =-\left(x^2+2x\left(y-1\right)+\left(y^2-2y+1\right)+\left(x^2-2x+1\right)-7\right)\\ =-\left(x^2+2x\left(y-1\right)+\left(y-1\right)^2+\left(x-1\right)^2-7\right)\\ =-\left(\left(x+y-1\right)^2+\left(x-1\right)^2-7\right)\\ =-\left(x+y-1\right)^2-\left(x-1\right)^2-7\)

\(\left(x+y-1\right)^2\ge0\\ \Rightarrow-\left(x+y-1\right)^2\le0\\ \left(x-1\right)^2\ge0\\ \Rightarrow-\left(x+y-1\right)^2-\left(x-1\right)^2\le0\\ \Rightarrow-\left(x+y-1\right)^2-\left(x-1\right)^2-7\le-7\)

Max A = -7 khi x=1 ; y=0

B) TT