Lời giải:

$A=(x^2+4y^2+4xy)+x^2+5-8x-12y$

$=(x+2y)^2-6(x+2y)+x^2+5-2x$

$=(x+2y)^2-6(x+2y)+9+(x^2-2x+1)-5$

$=(x+2y-3)^2+(x-1)^2-5\geq 0+0-5=-5$

Vậy $A_{\min}=-5$. Giá trị này đạt được khi $x+2y-3=x-1=0$

$\Leftrightarrow x=1; y=1$

Ta có : C = (x + 1).(x + 2).(x + 3).(x + 4)

=> C = [(x + 1).(x + 4)].[(x + 2).(x + 3)]

=> C = [x² + 5x + 4] . [x² + 5x + 6]

Đặt t = x² + 5x + 5

Khi đó t - 1 = x² + 5x + 4 , t + 1 = x² + 5x + 6  

Nên C = (t - 1)(t + 1) = t² - 1 = (x² + 5x + 5)² - 1

Mà (x² + 5x + 5)²​ \(\ge0\forall x\)

Do đó (x² + 5x + 5)² - 1​ \(\ge-1\forall x\)

Vậy GTNN của C là : 

H=\(x^6-2x^3+x^2-2x+2\)

\(=x^6+2x^5+3x^4+2x^2-2x^5-4x^4-6x^3-4x^2-4x+x^4+2x^3+3x^2+2x+2\)

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

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

\(=\left(x-1\right)^2\left(x^2+1\right)\left(x^2+2x+2\right)\)

\(=\left(x-1\right)^2\left(x^2+1\right)\left[\left(x+1\right)^2+1\right]\text{≥}0\)

Vì \(\left\{{}\begin{matrix}\left(x-1\right)^2\text{≥}0\\\left(x^2+1\right)\text{≥}1\\\left(x+1\right)^2+1\text{≥}1\end{matrix}\right.\)

⇒ MinH=0 ⇔ \(x=1\)

\(P=x^2+4xy+4y^2-4xy-4y^2+2x+3\)

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

dùng HĐT

\(=\left(4x^2-4xy+y^2\right)+\left(x^2+2x+1\right)+\left(3y^2-12y+12\right)-7\)

\(=\left(2x-y\right)^2+\left(x+1\right)^2+\left(\sqrt{3}y-\dfrac{12}{2\sqrt{3}}\right)^2-7\ge-7\)

\(\Rightarrow\) \(minD=-7\) khi \(\left\{{}\begin{matrix}\left(2x-y\right)^2=0\\\left(x+1\right)^2=0\\\left(\sqrt{3}y-\dfrac{12}{2\sqrt{3}}\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2x-y=0\\x+1=0\\\sqrt{3}y-\dfrac{12}{2\sqrt{3}}=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-1\\2\left(-1\right)-y=0\\\sqrt{3}y=\dfrac{12}{2\sqrt{3}}\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-1\\y=-2\\y=2\end{matrix}\right.\) \(\Rightarrow\) đề sai

\(A=-x^2-4y^2+2x-12y-10\)

\(A=-\left(x^2-2x+1\right)-\left(4y^2-12y+9\right)\)

\(A=-\left(x-1\right)^2-\left(2y+3\right)^2\)

Vậy\(A_{max}=0\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-\dfrac{3}{2}\end{matrix}\right.\)

Ta có: \(-x^2+2x-4y^2-12y-10\)

\(=-\left(x^2-2x+1+4y^2+12y+9\right)\)

\(=-\left(x-1\right)^2-\left(2y+3\right)^2\le0\forall x,y\)

Dấu '=' xảy ra khi \(\left(x,y\right)=\left(1;-\dfrac{3}{2}\right)\)