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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
\(C=-3x\left(3+x\right)-7=-9x-3x^2-7=-\left(3x^2+9x+7\right)=-3\left(x^2+3x+\frac{7}{3}\right)\)
=\(-3\left(x^2+2.\frac{3}{2}.x+\frac{9}{4}+\frac{1}{12}\right)=-3\left[\left(x+\frac{3}{2}\right)^2+\frac{1}{12}\right]=-3\left(x+\frac{3}{2}\right)^2-\frac{1}{4}\le-\frac{1}{4}\)
Dấu "=" xảy ra khi x=-3/2
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\(D=2xy-x^2-4y^2-8+2x+10y\)
\(=-\left(x^2+2xy-2x+4y^2-10y+8\right)\)
\(=-\left[x^2+2x\left(y-1\right)+4y^2-10y+8\right]\)
\(=-\left[x^2+2x\left(y-1\right)+\left(y^2-2y+1\right)+3y^2-8y+7\right]\)
\(=-\left[x^2+2x\left(y-1\right)+\left(y-1\right)^2+3\left(y^2-2.\frac{4}{3}.y+\frac{16}{9}\right)+\frac{5}{3}\right]\)
\(=-\left[\left(x+y-1\right)^2+3\left(y-\frac{4}{3}\right)^2+\frac{5}{3}\right]\)
\(=-\left(x+y-1\right)^2-3\left(y-\frac{4}{3}\right)^2-\frac{5}{3}\le-\frac{5}{3}\)
Dấu "=" xảy ra khi x=-1/3 và y=4/3
A=\(-x^2+2xy-4y^2+2x+8y-8=-\left(x^2-2xy+y^2-2x+1+2y\right)-\left(3y^2-6y+3\right)-4=-4-\left(x-y-1\right)^2-3\left(y-1\right)^2\le-4\)
=>Max A=-4<=>(x-y-1)2=0 và (y-1)2=0<=>x=2 y=1
\(A=-\left(x^2-2x\left(y+1\right)+\left(y+1\right)^2\right)-\left(4y^2-10y-5-\left(y+1\right)^2\right)\)
\(=-\left(x-y-1\right)^2-\left(3y^2-12y-6\right)\)
\(=-\left(x-y-1\right)^2-3\left(y-2\right)^2+18\le18\)
Max A=18 khi y=2; x=3