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\(Q=3xy\left(x+3y\right)-2xy\left(x+4y\right)-x^2\left(y-1\right)+y^2\left(1-x\right)+36\)\(\Leftrightarrow Q=3x^2y+9xy^2-2x^2y-8xy^2-x^2y+x^2+y^2-xy^2+36\)\(\Leftrightarrow Q=\left(3x^2y-2x^2y-x^2y\right)+\left(9xy^2-8xy^2-xy^2\right)+x^2+y^2+36\)\(\Leftrightarrow Q=x^2+y^2+36\ge36\forall x;y\)
Dấu " = " xảy ra
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=0\\y^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
Vậy Min Q là : \(36\Leftrightarrow x=y=0\)
\(D=x^2+5y^2-2xy+4y+3\)
\(=x^2-2xy+y^2+4y^2+4y+1+2\)
\(=\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)+2\)
\(=\left(x-y\right)^2+\left(2y+1\right)^2+2\)
Vì \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\forall x,y\\\left(2y+1\right)^2\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x-y\right)^2+\left(2y+1\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\forall x,y\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(2y+1\right)^2=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{1}{2}\)
Vậy \(D_{min}=2\Leftrightarrow x=y=-\dfrac{1}{2}\)
a) 25x2 - y2 + 4y - 4
= (5x)2 - (y - 2)2
= (5x + y - 2)(5x - y + 2)
b) a2 + b2 - x2 - y2 + 2ab - 2xy
= (a2 + 2ab + b2) - (x2 + 2xy + y2)
= (a + b)2 - (x + y)2
= (a + b + x + y)(a + b - x - y)
c) 5x2(x - 1) + 10xy(x - 1) - 5y2(1 - x)
= 5x2(x - 1) + 10xy(x - 1) + 5y2(x - 1)
= (x - 1)(5x2 + 10xy + 5y2)
= 5(x - 1)(x2 + 2xy + y2)
= 5(x -1)(x + y)2
d) x5 - x4y - xy4 + y5
= x4(x - y) - y4(x - y)
= (x - y)(x4 - y4)
= (x - y)(x2 - y2)(x2 + y2) = (x - y)2(x + y)(x2 + y2)
d= x2 + 5y2 + 2xy - 2y + 2005
d= x2 + 2xy + y2 + 4y2 - 2y + \(\frac{1}{4}+\)
d= ( x+ y )2 + ( 2y - \(\frac{1}{2}\))2 + \(\frac{8019}{4}\)\(\ge\)\(\frac{8019}{4}\)
dmin= \(\frac{8019}{4}khi\hept{\begin{cases}y=\frac{1}{4}\\x=-y=\frac{-1}{4}\end{cases}}\)