đạt gtnn là 17/4 khi x=căn bậc hai của 5 rồi chia cho 2 (2 không nằm trong dấu căn)

\(B=\sqrt{x}-x=-\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)

\(B_{max}=\dfrac{1}{4}\) khi \(\sqrt{x}-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{4}\)

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

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

\(A_{min}=-1\) 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 \(x=-\dfrac{1}{2}\)

GTNN:

Vì \(\sqrt{x}\ge0\Rightarrow3\sqrt{x}\ge0\Rightarrow P=3\sqrt{x}+1\ge1\)

Dấu bằng xảy ra <=> x=0

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

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

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