\(P=\left(x^2+4x+1\right)^2-12\left(x+2\right)^2+2093\)

\(P=\left(x^2+4x+1\right)^2-12\left(x^2+4x+1\right)+2093\)

\(P=\left(x^2+4x+1\right)^2-12\left(x^2+4x+1+3\right)+2093\)

Đặt: \(a=x^2+4x+1\)

\(\Rightarrow P=a^2-12\left(a+3\right)+2093\)

\(P=a^2-12a-36+2093\)

\(P=a^2-12a+2057\)

\(P=a^2-12a+36+2021\)

\(P=\left(a^2-2\cdot6\cdot a+6^2\right)+2021\)

\(P=\left(a-6\right)^2+2021\)

Ta có: \(\left(a-6\right)^2\ge0\forall a\)

\(\Rightarrow P=\left(t-6\right)^2+2021\ge2021\)

\(\Rightarrow P\ge2021\Rightarrow P_{min}=2021\)

Dấu "=" xảy ra: \(\left(t-6\right)^2=0\Leftrightarrow t-6=0\Leftrightarrow t=6\)

Vậy: \(P_{min}=2021\) khi \(t=6\)

Mà: \(t=6\Rightarrow x^2+4x+1=6\)

\(\Leftrightarrow x^2+4x+1-6=0\)

\(\Leftrightarrow x^2+4x-5=0\)

\(\Leftrightarrow x^2-x+5x-5=0\)

\(\Leftrightarrow x\left(x-1\right)+5\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

Vậy: \(P_{min}=2021\) khi \(\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

\(P=\left(x^2+4x+1\right)^2-12\left(x+2\right)^2+2093\\ P=\left(x^2+4x+1\right)^2-12\left(x^2+4x+4\right)+2093\\ P=\left(x^2+4x+1\right)^2-2\left(x^2+4x+1\right).6-36+2093\\ P=\left(x^2+4x+1\right)^2-2\left(x^2+4x+1\right).6+36+2021\\ P=\left(x^2+4x-5\right)^2+2021\ge2021\)

Dấu "=" xảy ra tương đương với \(\left\{{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

\(A=-\dfrac{4}{x^2-4x+10}\\ =-\dfrac{4}{\left(x^2-2.x.2+4+6\right)}\\ =-\dfrac{4}{\left(x-2\right)^2+6}\)

\(\left(x-2\right)^2\ge0\\ \Rightarrow\left(x-2\right)^2+6\ge6\\ \Rightarrow\dfrac{4}{\left(x-2\right)^2+6}\le\dfrac{2}{3}\\ \Rightarrow A=-\dfrac{4}{\left(x-2\right)^2+6}\ge-\dfrac{2}{3}\)

Min A=-2/3 khi x=2

\(C=\dfrac{2}{x^2+4x+5}=\dfrac{2}{\left(x+2\right)^2+1}\)

Vì \(\left(x+2\right)^2\ge0\Rightarrow\left(x+2\right)^2+1\ge1\)

\(\Rightarrow C\le2\)

Dấu ''='' xảy ra \(\Leftrightarrow x=-2\)

Vậy Min C = 2 kjhi x = -2

a) Ta có: \(Q=-x^2-y^2+4x-4y+2=-\left(x^2+y^2-4x+4y-2\right)\)

\(=-\left(x^2-4x+4+y^2+4y+4\right)+10\)

\(=-\left[\left(x-2\right)^2+\left(y+2\right)^2\right]+10\le10\forall x,y\)

Vậy Max_Q=10 khi x=2, y=-2

b) +Ta có: \(A=-x^2-6x+5=-\left(x^2+6x-5\right)=-\left(x^2+6x+9-14\right)\)

\(=-\left(x^2+6x+9\right)+14=-\left(x+3\right)^2+14\le14\forall x\)

Vậy Max_A=14 khi x=-3

+Ta có: \(B=-4x^2-9y^2-4x+6y+3=-\left(4x^2+9y^2+4x-6y-3\right)\)

\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)

\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2\right]+5\le5\forall x,y\)

Vậy Max_B=5 khi x=-1/2, y=1/3

c) Ta có: \(P=x^2+y^2-2x+6y+12=x^2-2x+1+y^2+6y+9+2\)

\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\forall x,y\)

Vậy Min_P=2 khi x=1, y=-3