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\(F=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\dfrac{x+y}{xy}\cdot\dfrac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}+\dfrac{2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)^2}\right]\)

\(=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\dfrac{x+y+2\sqrt{xy}}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right]\)

\(=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}\cdot xy=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)

\(\left(1\right)=\dfrac{y}{x\left(2x-y\right)}-\dfrac{4x}{y\left(2x-y\right)}=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(y-2x\right)\left(y+2x\right)}{xy\left(y-2x\right)}=\dfrac{-y-2x}{xy}\\ \left(2\right)=\dfrac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x+6}{\left(x+2\right)^2}\\ \left(3\right)=\dfrac{4\left(x+2\right)}{\left(x+2\right)\left(4x+7\right)}=\dfrac{4}{4x+7}\\ \left(4\right)=\dfrac{4x^2+15x+4+4x+7+1}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}=\dfrac{4x^2+19x+12}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}\)

a: \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+2\right)\left(y+3\right)=xy+100\\\left(x-2\right)\left(y-2\right)=xy-64\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+2y=94\\-2x-2y=-68\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=26\\y=8\end{matrix}\right.\)

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

c: \(\Leftrightarrow\left\{{}\begin{matrix}xy-2x=xy-4x+2y-8\\2xy+7x-6y-21=2xy+6x-7y-21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-2y=-8\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=2\end{matrix}\right.\)

a: \(\frac{\sqrt{x}-\sqrt{y}}{xy\cdot\sqrt{xy}}:\left(\frac{1}{x}+\frac{1}{y}\right)\cdot\frac{1}{x+y+2\sqrt{xy}}\)

\(=\frac{\sqrt{x}-\sqrt{y}}{xy\cdot\sqrt{xy}}:\frac{x+y}{xy}\cdot\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}\)

\(=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)^2}\)

\(\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\)

\(=\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}\cdot\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)

\(=\frac{2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)^2}=\frac{2\left(x+y\right)}{\sqrt{xy}\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)^2}\)

Ta có: \(C=\frac{\sqrt{x}-\sqrt{y}}{xy\cdot\sqrt{xy}}:\left(\frac{1}{x}+\frac{1}{y}\right)\cdot\frac{1}{x+y+2\sqrt{xy}}+\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\)

\(=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2\left(x+y\right)}{\sqrt{xy}\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)^2}=\frac{2\left(x+y\right)+\sqrt{x}-\sqrt{y}}{\sqrt{xy}\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)^2}\)

b: \(\left(\sqrt{x}+\sqrt{y}\right)^2=\left(\sqrt{2-\sqrt3}+\sqrt{2+\sqrt3}\right)^2\)

\(=2-\sqrt3+2+\sqrt3+2\cdot\sqrt{\left(2-\sqrt3\right)\left(2+\sqrt3\right)}=4+2=6\)

\(\sqrt{xy}=\sqrt{\left(2+\sqrt3\right)\left(2-\sqrt3\right)}=\sqrt{4-3}=1\)

\(x+y=2+\sqrt3+2-\sqrt3=4\)

\(\sqrt{x}-\sqrt{y}=\sqrt{2-\sqrt3}-\sqrt{2+\sqrt3}\)

\(=\frac{1}{\sqrt2}\left(\sqrt{4-2\sqrt3}-\sqrt{4+2\sqrt3}\right)\)

\(=\frac{1}{\sqrt2}\left(\sqrt{\left(\sqrt3-1\right)^2}-\sqrt{\left(\sqrt3+1\right)^2}\right)=\frac{1}{\sqrt2}\left(\sqrt3-1-\sqrt3-1\right)=-\frac{2}{\sqrt2}=-\sqrt2\)

Ta có: \(C=\frac{2\left(x+y\right)+\sqrt{x}-\sqrt{y}}{\sqrt{xy}\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)^2}\)

\(=\frac{2\cdot4-\sqrt2}{1\cdot4\cdot6}=\frac{8-\sqrt2}{24}\)

a: \(=\dfrac{1}{x-y}-\dfrac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\dfrac{x-y}{x^2+xy+y^2}\)

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

\(=\dfrac{2x^2-4xy+2y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{2\left(x-y\right)}{x^2+xy+y^2}\)

d: \(=\dfrac{x^3-1}{x-1}-\dfrac{x^2-1}{x+1}\)

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

`@ x+y+z=1`.

`<=>` \(\left\{{}\begin{matrix}x=1-y-z\\y=1-z-x\\z=1-x-y\end{matrix}\right.\)

`P=(x+y)^2/(xy+1-x-y).(y+z)^2/(yz-y-z+1).(x+z)^2/(xy-x-y+1)`.

`<=> ((1-z)^2(1-y)^2(1-x)^2)/((1-x)(1-y)(1-y)(1-z)(1-z)(1-x).`

`=1.`

Vậy `P` không phụ thuộc vào giá trị của biến.

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}\left(xy+3x+2y+6\right)=\dfrac{1}{2}xy+56\\\dfrac{1}{2}\left(xy-2x-2y+4\right)=\dfrac{1}{2}xy-32\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+2y+6=112\\-2x-2y+4=-64\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+2y=106\\-2x-2y=-68\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+2y=106\\x=38\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=38\\y=-4\end{matrix}\right.\)

Áp dụng bất đẳng thức AM - GM:

\(P\ge3\sqrt[3]{\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}\).

Áp dụng bất đẳng thức AM - GM ta có:

\(xy+1=xy+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}\ge5\sqrt[5]{\dfrac{xy}{4^4}}\).

Tương tự: \(yz+1\ge5\sqrt[5]{\dfrac{yz}{4^4}};zx+1\ge5\sqrt[5]{\dfrac{zx}{4^4}}\).

Do đó \(\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)\ge125\sqrt[5]{\dfrac{\left(xyz\right)^2}{4^{12}}}\)

\(\Rightarrow\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}\ge125\sqrt[5]{\dfrac{1}{4^{12}\left(xyz\right)^3}}\).

Mà \(xyz\le\dfrac{\left(x+y+z\right)^3}{27}=\dfrac{1}{8}\)

Nên \(\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}\ge125\sqrt[5]{\dfrac{8^3}{4^{12}}}=125\sqrt[5]{\dfrac{1}{2^{15}}}=\dfrac{125}{8}\)

\(\Rightarrow P\ge\dfrac{15}{2}\).

Vậy...

Áp dụng bất đẳng thức AM - GM:

P≥33√(xy+1)(yz+1)(zx+1)xyz.

Áp dụng bất đẳng thức AM - GM ta có:

xy+1=xy+14+14+14+14≥55√xy44.

Tương tự: yz+1≥55√yz44;zx+1≥55√zx44.

Do đó (xy+1)(yz+1)(zx+1)≥1255√(xyz)2412

⇒(xy+1)(yz+1)(zx+1)xyz≥1255√1412(xyz)3.

Mà xyz≤(x+y+z)327=18

Nên  (xy+1)(yz+1)(zx+1)xyz≥1255√83412=1255√1215=1258 

⇒P≥152.