\(a,=-\left(x^2-2x+1\right)-3=-\left(x-1\right)^2-3\le-3\)

Dấu \("="\Leftrightarrow x=1\)

\(b,=-\left(x^2+4x+4\right)+4=-\left(x+2\right)^2+4\le4\)

Dấu \("="\Leftrightarrow x=-2\)

\(c,=-\left(9x^2-24x+16\right)-2=-\left(3x-4\right)^2-2\le-2\)

Dấu \("="\Leftrightarrow x=\dfrac{4}{3}\)

\(d,=-\left(x^2-4x+4\right)+3=-\left(x-2\right)^2+3\le3\)

Dấu \("="\Leftrightarrow x=2\)

\(x^2+y^2+xy=3\)

Có \(x^2+y^2\ge2xy\) \(\Rightarrow3=x^2+y^2+xy\ge2xy+xy\) \(\Leftrightarrow xy\le1\)

\(x^2+y^2\ge-2xy\) \(\Rightarrow3=x^2+y^2+xy\ge-2xy+xy\) \(\Leftrightarrow-3\le xy\) 

Đặt A= \(x^2+y^2-xy=\left(3-xy\right)-xy=3-2xy\)

mà \(-3\le xy\le1\) \(\Rightarrow9\ge3-2xy\ge1\)

=> minA=1 <=> \(\left\{{}\begin{matrix}xy=1\\x=y\end{matrix}\right.\) <=>x=y=1

maxA=9 <=>\(\left\{{}\begin{matrix}xy=-3\\x=-y\end{matrix}\right.\) <=>\(\left(x;y\right)=\left(\sqrt{3};-\sqrt{3}\right);\left(-\sqrt{3};\sqrt{3}\right)\)

Đặt \(P=x^2+y^2-xy\)

\(\Rightarrow\dfrac{P}{3}=\dfrac{x^2+y^2-xy}{3}=\dfrac{x^2+y^2-xy}{x^2+y^2+xy}\)

\(\dfrac{P}{3}=\dfrac{3x^2+3y^2-3xy}{3\left(x^2+y^2+xy\right)}=\dfrac{x^2+y^2+xy+2\left(x^2+y^2-2xy\right)}{3\left(x^2+y^2+xy\right)}\)

\(\dfrac{P}{3}=\dfrac{1}{3}+\dfrac{2\left(x-y\right)^2}{3\left(x^2+y^2+xy\right)}\ge\dfrac{1}{3}\Rightarrow P\ge1\)

\(P_{min}=1\) khi \(x=y=1\)

\(\dfrac{P}{3}=\dfrac{x^2+y^2-xy}{x^2+y^2+xy}=\dfrac{3\left(x^2+y^2+xy\right)-2\left(x^2+y^2+2xy\right)}{x^2+y^2+xy}=3-\dfrac{2\left(x+y\right)^2}{x^2+y^2+xy}\le3\)

\(\Rightarrow P\le9\)

\(P_{max}=9\) khi \(\left(x;y\right)=\left(\sqrt{3};-\sqrt{3}\right);\left(-\sqrt{3};\sqrt{3}\right)\)

a) Áp dụng bất đẳng thức Cosi ta có :

\(x^2+1\geq 2x\\ 4y^2+1\geq 4y\\ 9z^2+1\geq 6z\)

Suy ra \(S\leq 6\)

Dấu = xảy ra khi \(x=1;y=\frac{1}{2}; z=\frac{1}{3}\)

a) x² +x +1 = x² + x + 1/4 + 3/4 =(x+1/2)² + 3/4

=> GTNN a) =3/4 khi x=-1/2

b) 4x² +4x -5 = 4x² + 4x +1 -6 = (2x+1)²-6

=> GTNN b) = -6 khi x=-1/2

c) (x-3)(x+5) +4 = x²+2x -11 = x²+2x +1-12=(x+1)²-12

GTNN c) =12 khi x=-1 

d) x²-4x+y²-8y+6=x²-4x+4+y²-8y+16-14=(x-2)²+(y-4)²-14

GTNN d) =-14 khi x=2 , y=4

\(a,=\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Dấu \("="\Leftrightarrow x=-\dfrac{1}{2}\)

\(b,=\left(4x^2+4x+1\right)-6=\left(2x+1\right)^2-6\ge-6\)

Dấu \("="\Leftrightarrow x=-\dfrac{1}{2}\)

\(c,=x^2+2x-15+4=\left(x+1\right)^2-12\ge-12\)

Dấu \("="\Leftrightarrow x=-1\)

\(d,=\left(x^2-4x+4\right)+\left(y^2-8y+16\right)-14=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)

\(A=-\left(x^2-4x-3\right)=-\left(x^2-4x+4-7\right)=-\left(x-2\right)^2+7\le7\)

\(A_{max}=7\Leftrightarrow x=2\)