Bài 2:
a: \(A=x-x^2\)
\(=-x^2+x-\dfrac{1}{4}+\dfrac{1}{4}\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}< =\dfrac{1}{4}\forall x\)
Dấu '=' xảy ra khi \(x-\dfrac{1}{2}=0\)
=>\(x=\dfrac{1}{2}\)
b: \(B=-2x^2+2x-5\)
\(=-2\left(x^2-x+\dfrac{5}{2}\right)\)
\(=-2\left(x^2-x+\dfrac{1}{4}+\dfrac{9}{4}\right)\)
\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}< =-\dfrac{9}{2}\forall x\)
Dấu '=' xảy ra khi \(x-\dfrac{1}{2}=0\)
=>\(x=\dfrac{1}{2}\)
c: \(\left(x+3\right)^2>=0\forall x;\left|y-1\right|>=0\forall y\)
Do đó: \(\left(x+3\right)^2+\left|y-1\right|>=0\forall x,y\)
=>\(\left(x+3\right)^2+\left|y-1\right|+5>=5\forall x,y\)
=>\(C=\dfrac{2005}{\left(x+3\right)^2+\left|y-1\right|+5}< =\dfrac{2005}{5}=401\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+3=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)
d: \(D=-4x^2-12x+10\)
\(=-4x^2-12x-9+19\)
\(=-\left(4x^2+12x+9\right)+19=-\left(2x+3\right)^2+19< =19\forall x\)
Dấu '=' xảy ra khi 2x+3=0
=>2x=-3
=>\(x=-\dfrac{3}{2}\)
e: \(E=2x-x^2-2\)
\(=-x^2+2x-1-1\)
\(=-\left(x^2-2x+1\right)-1=-\left(x-1\right)^2-1< =-1\forall x\)
Dấu '=' xảy ra khi x-1=0
=>x=1
f: \(F=-x^2-y^2+6x-4y+7\)
\(=-x^2+6x-9-y^2-4y-4+20\)
\(=-\left(x-3\right)^2-\left(y+2\right)^2+20< =20\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-3=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-2\end{matrix}\right.\)
