\(=2xy+5x+4y^2+10y\)

\(=x\left(2y+5\right)+2y\left(2y+5\right)\)

\(=\left(x+2y\right)\left(2y+5\right)\)

\(=2x\left(x+2y\right)+5\left(x+2y\right)=\left(x+2y\right)\left(2x+5\right)\)

Giải:

a) x² - 2xy + y² - 25 = (x - y)² - 5² = (x - y - 5)(x - y + 5)

b) x² - 10x - y² + 25 = (x² - 10x + 25) - y² = (x - 5)² - y² = (x - 5 - y)(x - 5 + y)

c) 3x + 3y + zx + zy = (3x + 3y) + (zx + zy) = 3(x + y) + z(x + y) = (x + y)(3 + z)

d) x² - 4y² + 5x +10y = (x² - 4y²) + (5x +10y) = (x - 2y)(x + 2y) + 5(x + 2y) = (x + 2y)(x - 2y + 5)

\(D=-5\left(x^2+\dfrac{4}{5}x+\dfrac{1}{5}\right)\)

\(=-5\left(x^2+2\cdot x\cdot\dfrac{2}{5}+\dfrac{4}{25}+\dfrac{1}{25}\right)\)

\(=-5\left(x+\dfrac{2}{5}\right)^2-\dfrac{1}{5}< =-\dfrac{1}{5}\)

Dấu = xảy ra khi x=-2/5

Ta có \(A=-x^2+2xy-4y^2+2x+10y-3\) 

\(A=-x^2+2\left(y+1\right)x-4y^2+10y-3\)

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

\(A=-\left[x-\left(y+1\right)\right]^2-3\left(y^2-4y+4\right)+10\)

\(A=-\left(x-\left(y+1\right)\right)^2-3\left(y-2\right)^2+10\) \(\le10\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=y+1\\y-2=0\end{matrix}\right.\Leftrightarrow\left(x,y\right)=\left(3,2\right)\)

Vậy \(max_A=10\)

?

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

\(=-x^2+2xy-y^2+2x-2y-1-3y^2+12y-12+10\)

\(=-\left(x^2-2xy+y^2-2x+2y+1\right)-3\left(y^2-4y+4\right)+10\)

\(=-\left(x-y-1\right)^2-3\left(y-2\right)^2+10< =10\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-y-1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=y+1=3\end{matrix}\right.\)

\(B=-4x^2-5y^2+8xy+10y+12\)

\(=-4x^2+8xy-4y^2-y^2+10y-25+37\)

\(=-4\left(x^2-2xy+y^2\right)-\left(y^2-10y+25\right)+37\)

\(=-4\left(x-y\right)^2-\left(y-5\right)^2+37< =37\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-y=0\\y-5=0\end{matrix}\right.\)

=>x=y=5

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

\(=-\left(x^2-2xy+4y^2-2x-10y+8\right)\)

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

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

Dấu"=" xảy ra  <=>  \(\hept{\begin{cases}x-y-1=0\\y-2=0\end{cases}}\) <=>  \(\hept{\begin{cases}x=3\\y=2\end{cases}}\)

Vậy MAX  \(A=5\)khi  \(x=3;\)\(y=2\)

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

\(=-\left(x^2+2xy-y^2\right)+2x-2y-1-3y^2+12y-12+10\)

\(=-\left(x-y\right)^2+2\left(x-y\right)-1-3\left(y^2-4y+4\right)+10\)

\(=-\left(x-y-1\right)^2-3\left(y-2\right)^2+10\le10\forall x;y\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-y-1=0\\y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=3\\y=2\end{cases}}}\)

Vậy \(C_{max}=10\) tại x = 3; y = 2