a) \(xy+2x+3y=-6\)

\(\Rightarrow x\left(y+2\right)+3y+6=0\)

\(\Rightarrow x\left(y+2\right)+3\left(y+2\right)=0\)

\(\Rightarrow\left(x+3\right)\left(y+2\right)=0\)

\(\Rightarrow\left[\begin{matrix}x+3=0\\y+2=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)

Vậy \(x=-3;y=-2\)

\(xy+2x+3y=-6\)

\(\Leftrightarrow xy+2x+3y+6=0\)

\(\Leftrightarrow y\left(x+3\right)+\text{2}\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(y+2\right)=0\)

\(\Leftrightarrow\left\{\begin{matrix}x+3=0\\y+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)

Vậy \(\left\{\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)

\(xy+2x+3y=-6\)

\(\Leftrightarrow x\left(y+2\right)+3y+6=-6+6\)

\(\Leftrightarrow x\left(y+2\right)+3\left(y+2\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(y+2\right)=0\)

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

Đặt \(\hept{\begin{cases}\sqrt{2x+3}=a\left(a>0\right)\\\sqrt{y}=b\left(b\ge0\right)\end{cases}}\)

Thì ta có

\(\frac{b^2}{a^2}=\frac{a+1}{b+1}\)

\(\Leftrightarrow b^3+b^2=a^3+a^2\)

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

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

Mà \(\left(b^2+ab+a^2+b+a\right)>0\)

\(\Rightarrow a=b\)

\(\Rightarrow2x+3=y\)

Thế vào Q ta được 

\(Q=2x^2-5x-12=\left(2x^2-\frac{2x\times\sqrt{2}\times5}{2\sqrt{2}}+\frac{25}{8}\right)-\frac{121}{8}\)

\(=\left(\sqrt{2}x-\frac{5}{2\sqrt{2}}\right)^2-\frac{121}{8}\ge\frac{-121}{8}\)

\(\Leftrightarrow xy-2x+3y-6=11\)

\(\Leftrightarrow x\left(y-2\right)+3\left(y-2\right)=11\)

\(\Leftrightarrow\left(x+3\right)\left(y-2\right)=11\)

Bảng giá trị:

Điều kiện \(x\ne\pm3;y\ne-2\):

 \(P=\frac{2x+3y}{xy+2x-3y-6}-\frac{6-xy}{xy+2x+3y+6}-\frac{x^2+9}{x^2-9}.\)

=> \(P=\frac{2x+3y}{\left(y+2\right)\left(x-3\right)}-\frac{6-xy}{\left(y+2\right)\left(x+3\right)}-\frac{x^2+9}{\left(x-3\right)\left(x+3\right)}\)

\(P=\frac{\left(2x+3y\right)\left(x+3\right)-\left(6-xy\right)\left(x-3\right)-\left(x^2+9\right)\left(y+2\right)}{\left(y+2\right)\left(x-3\right)\left(x+3\right)}\)

\(P=\frac{2x^2+3xy+6x+9y-6x+x^2y+18-3xy-x^2y-9y-2x^2-18}{\left(y+2\right)\left(x-3\right)\left(x+3\right)}\)

\(P=\frac{0}{\left(y+2\right)\left(x-3\right)\left(x+3\right)}=0\)

=> P=0 (với mọi x khác 3, -3 và y khác -2)

`2x^2+3y^2+4z^2-2(x+y+z)+2`

`=2x^2-2x+1/2+3y^2-2y+1/3+4z^2-2z+1/4+11/12`

`=2(x-1/2)^2+3(y-1/3)^2+4(z-1/4)^2+11/12>=11/12`

Dấu "=" xảy ra khi \(\begin{cases}x=\dfrac12\\y=\dfrac13\\z=\dfrac14\\\end{cases}\)

x= 75 ; y = 50 ; z = 30