Lời giải:

\(A=3x^2+11y^2-2xy-2x+6y-1\)

\(\Leftrightarrow A=\left(x^2+y^2+\frac{1}{4}-2xy-x+y\right)+2\left(x^2-\frac{1}{2}x+\frac{1}{16}\right)+10\left(y^2+\frac{1}{2}y+\frac{1}{16}\right)-2\)

\(\Leftrightarrow A=\left(x-y-\frac{1}{2}\right)^2+2\left(x-\frac{1}{4}\right)^2+10\left(y+\frac{1}{4}\right)^2-2\)

Thấy rằng \(\hept{\begin{cases}\left(x-y-\frac{1}{2}\right)^2\ge0\\\left(x-\frac{1}{4}\right)^2\ge0\\\left(y+\frac{1}{4}\right)^2\ge0\end{cases}}\Rightarrow A\ge-2\)

Vậy \(A_{min}=-2\Leftrightarrow\hept{\begin{cases}x-y-\frac{1}{2}=0\\x-\frac{1}{4}=0\\y+\frac{1}{4}=0\end{cases}\Leftrightarrow x=\frac{1}{4};y=\frac{-1}{4}}\)

\(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)

\(=\frac{1-a\sqrt{a}+\sqrt{a}-a}{1-\sqrt{a}}.\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)

\(=\left(1-a\sqrt{a}+\sqrt{a}-a\right)\frac{1-\sqrt{a}}{\left(1-a\right)^2}\)

\(=\frac{1-a\sqrt{a}+\sqrt{a}-a-\sqrt{a}+a.\left(\sqrt{a}\right)^2-\left(\sqrt{a}\right)^2+a\sqrt{a}}{\left(1-a\right)^2}\)

\(=\frac{a^2-2a+1}{\left(1-a\right)^2}=\frac{\left(a-1\right)^2}{\left(1-a\right)^2}\)

\(=\left(\frac{a-1}{1-a}\right)^2=\left(-1\right)^2=1=VP\left(ĐPCM\right)\)

\(PT=\sqrt{x-2}+\sqrt{4-x}\le\sqrt{2\left(x-2+4-x\right)}=4\)

\(PT=\left(x-3\right)^2+2\ge2\)

P/s: Ko chắc

Ta có \(x^2+6x+11=\left(x+6\right)\sqrt{x^2+11}\Leftrightarrow\left(x^2+11\right)+6\left(x+6\right)-36=\left(x+6\right)\sqrt{x^2+11}\)

Đặt \(\hept{\begin{cases}x+6=a\\\sqrt{x^2+11}=b\end{cases}}\left(b>0\right)\)

Phương trình trở thành:        \(b^2+6a-36=ab\)

\(\Leftrightarrow b^2-36+6a-ab=0\Leftrightarrow\left(b-6\right)\left(b+6\right)+a\left(6-b\right)=0\)

\(\Leftrightarrow\left(b-6\right)\left(b+6-a\right)=0\Leftrightarrow\orbr{\begin{cases}b=6\\b-a+6=0\end{cases}}\)

TH1: \(b=6\Leftrightarrow\sqrt{x^2+11}=6\Leftrightarrow x^2+11=36\Leftrightarrow x^2=25\Leftrightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)

TH2: \(b-a+6=0\Leftrightarrow b=a-6\)

Trở về ẩn x, ta có:   \(\sqrt{x^2+11}=\left(x+6\right)-6\Leftrightarrow\sqrt{x^2+11}=x\left(x>0\right)\)

\(\Leftrightarrow x^2+11=x^2\) (Vô lý)

Vậy phương trình có 2 nghiệm x = 5 hoặc x = - 5.

\(\left(y+2\right)x^2+1=y^2\Leftrightarrow x^2y+2x^2+1-y^2=0\Leftrightarrow\)\(x^2y+2x^2+4-y^2-3=0\Leftrightarrow x^2\left(y+2\right)-\left(y^2-4\right)=3\)\(\Leftrightarrow x^2\left(y+2\right)-\left(y+2\right)\left(y-2\right)=3\)

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

Ta có bảng:

Vậy ta tìm được cặp (x ; y) = (0 ; 1) và (0; -1).

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

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

+, Trường hợp: \(\hept{\begin{cases}y+2=3\\x^2+2-x=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\y=1\end{cases}}\)

+, Trường hợp: \(\hept{\begin{cases}y+2=1\\x^2+2-x=3\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=0\\y=-1\end{cases}}\)