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

\(A_{min}=-1\) khi \(x=-2\)

\(A=\dfrac{4x^2+4-4x^2+4x-1}{x^2+1}=4-\dfrac{\left(2x-1\right)^2}{x^2+1}\le4\)

\(A_{max}=4\) khi \(x=\dfrac{1}{2}\)

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

\(A_{min}=-1\) khi \(x=2\)

\(A=\dfrac{4x^2+4-4x^2-4x-1}{x^2+1}=4-\dfrac{\left(2x+1\right)^2}{x^2+1}\le4\)

\(A_{max}=4\) khi \(x=-\dfrac{1}{2}\)

a: \(A\ge1\forall x\)

Dấu '=' xảy ra khi x=3/4

\(x^2-4x+7\) 

⇔ \(\left(x^2-4x+4\right)+3\)

⇔  \(\left(x-2\right)^2+3\)

Vì \(\left(x-2\right)^2\ge0\) ⇒ \(\left(x-2\right)^2+3\ge3\)

Vậy GTNN của A là 3 khi x =2  

\(x^2-4x+7\)

\(=x^2-4x+4+3\)

\(=\left(x-2\right)^2+3\ge3\forall x\)

Dấu '=' xảy ra khi x=2

Ta có : -x²-4x+9

=-x²-4x-4+13

=-(x²+4x+4)+13

=-(x+2)²+13

=13-(x+2)²

\(\Rightarrow\)(x+2)²\(\ge\)0

Ma: 13>0 \(\Leftrightarrow\)(x+2)²\(\le\)13

Vay GTLN la 13

Dau "=" xay ra khi : x+2=0

                                 x=-2

-x^2-4x+9=-(x^2+4x+4-13)=-(x+2)^2+13 

ta co -(x+2)^2 nho hon hoac bang 0

               13 lon hon 0

nen bt tren se nho hon hoac bang 13

 dau = xay ra <=> x+2=0=>x=-2

vay min bt =13 tai x=-2

\(T=-2\left(x^2+y^2+1-2xy+2x-2y\right)-2y^2+8y+2004\)

\(T=-2\left(x-y+1\right)^2-2\left(y-2\right)^2+2012\le2012\)

\(T_{max}=2012\) khi \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

x² + 4x + 4 = (x + 2)²  < hoặc = 0

max A = 0 <=> x = -2