a: \(=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y}{x-y}\cdot\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

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

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

\(=\sqrt{x}-\sqrt{y}-\sqrt{x}+\sqrt{y}=0\)

c: \(=\dfrac{x-1-4\sqrt{x}+\sqrt{x}+1}{x-1}\cdot\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x-3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\)

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

\(B=\dfrac{\sqrt{x}-\sqrt{y}}{x\sqrt{x}-y\sqrt{y}}=\dfrac{1}{x+\sqrt{xy}+y}\)

\(C=\dfrac{3\sqrt{3}+x\sqrt{x}}{3-\sqrt{3x}+x}=\sqrt{x}+\sqrt{3}\)

\(D=\dfrac{x+\sqrt{5x}+5}{x\sqrt{x}-5\sqrt{5}}=\dfrac{1}{\sqrt{x}-\sqrt{5}}\)

`[\sqrt{27}-\sqrt{15}]/[3-\sqrt{5}]+4/[2+\sqrt{3}]-6/\sqrt{3}`

`=[\sqrt{3}(3-\sqrt{5})]/[3-\sqrt{5}]+[4(2-\sqrt{3})]/[4-3]-[2\sqrt{3}.\sqrt{3}]/\sqrt{3}`

`=\sqrt{3}+8-4\sqrt{3}-2\sqrt{3}`

`=8-5\sqrt{3}`

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`[x-y]/[\sqrt{x}+\sqrt{y}]-[x\sqrt{y}+y\sqrt{x}]/\sqrt{xy}`    `ĐK:  x,y > 0`

`=[(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})]/[\sqrt{x}+\sqrt{y}]-[\sqrt{xy}(\sqrt{x}+\sqrt{y})]/\sqrt{xy}`

`=\sqrt{x}-\sqrt{y}-\sqrt{x}-\sqrt{y}`

`=-2\sqrt{y}`

a) Ta có: \(P=\left[\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)\cdot\dfrac{2}{\sqrt{x}+\sqrt{y}}+\dfrac{1}{x}+\dfrac{1}{y}\right]:\dfrac{\sqrt{x^3}+y\sqrt{x}+x\sqrt{y}+\sqrt{y^3}}{\sqrt{x^3y}+\sqrt{xy^3}}\)

\(=\left(\dfrac{2}{\sqrt{xy}}+\dfrac{1}{x}+\dfrac{1}{y}\right):\dfrac{x\sqrt{x}+y\sqrt{x}+x\sqrt{y}+y\sqrt{y}}{x\sqrt{xy}+y\sqrt{xy}}\)

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

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

\(=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)

a) Đk:\(x>0;y>0\)

\(P=\left[\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}.\sqrt{y}}.\dfrac{2}{\sqrt{x}+\sqrt{y}}+\dfrac{1}{x}+\dfrac{1}{y}\right]:\dfrac{x\left(\sqrt{x}+\sqrt{y}\right)+y\left(\sqrt{x}+\sqrt{y}\right)}{x\sqrt{xy}+y\sqrt{xy}}\)

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

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

b) \(xy=16\Leftrightarrow x=\dfrac{16}{y}\)

\(P=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}=\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}=\dfrac{1}{\sqrt{\dfrac{16}{y}}}+\dfrac{1}{\sqrt{y}}=\dfrac{\sqrt{y}}{4}+\dfrac{1}{\sqrt{y}}\)

Áp dụng AM-GM có:

\(\dfrac{\sqrt{y}}{4}+\dfrac{1}{\sqrt{y}}\ge2\sqrt{\dfrac{\sqrt{y}}{4}.\dfrac{1}{\sqrt{y}}}=1\)

\(\Rightarrow P\ge1\)

Dấu "=" xảy ra khi \(y=4\Rightarrow x=4\)

Vậy x=y=4 thì P đạt GTNN là 1

Bổ sung giả thuyết x ,y \(\ge0\)

Do giả thiết x ,y \(\ge0\)

\(\sqrt{x}+\sqrt{y}\) =1
nên:
xy (x+y )\(^2\)\(\le\) \(\dfrac{1}{64}\)
<=> 64 xy (x + y )\(^2\) \(\le\)1
<=> 64 xy ( x + y)\(^2\)\(\le\)(\(\sqrt{x}+\sqrt{y}\))\(^8\)
<=> 64 xy ( x + y )\(^2\) < \((x+2\sqrt{xy}+y)^4\)
Áp dụng bất đẳng thức Cauchy cho 2 số không âm x + y và \(2\sqrt{xy}\)
ta có ;
x + y + 2\(\sqrt{xy}\) \(\ge\) \(2\sqrt{x+y}2\sqrt{xy}\)
=> ( x + y +2\(\sqrt{xy}\)) \(^4\)\(\ge\) (\(2\sqrt{x+y}2\sqrt{xy}\) )\(^4\)= 64 xy (x + y)\(^2\)
=> ĐIỀU PHẢI CHỨNG MINH
Dấu bằng xảy ra <=> x + y = \(2\sqrt{xy}\)
<=> x = y = \(\dfrac{1}{4}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\\\sqrt{y}=b\end{matrix}\right.\), ta có:

\(A=\left[\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\times\dfrac{2}{a+b}+\dfrac{1}{a^2}+\dfrac{1}{b^2}\right]\)\(\times\dfrac{a^3+ab^2+a^2b+b^3}{ab^3+a^3b}\)

\(=\left(\dfrac{b+a}{ab}\times\dfrac{2}{a+b}+\dfrac{b^2+a^2}{a^2b^2}\right)\)\(\times\dfrac{a^2\left(a+b\right)+b^2\left(a+b\right)}{ab\left(a^2+b^2\right)}\)

\(=\dfrac{2ab+b^2+a^2}{a^2b^2}\times\dfrac{\left(a+b\right)\left(a^2+b^2\right)}{ab\left(b^2+a^2\right)}\)

\(=\dfrac{\left(a+b\right)^3}{a^3b^3}\)

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

a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{3}=\dfrac{13}{6}\sqrt{6}-2\sqrt{3}\)

b: \(VT=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\cdot\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)^2\)

c: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)

\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)

\(\dfrac{5\left(4+\sqrt{11}\right)}{\left(4+\sqrt{11}\right)\left(4-\sqrt{11}\right)}+\dfrac{3-\sqrt{7}}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}-\dfrac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\dfrac{\sqrt{7}-5}{2}\)\(=\dfrac{\left(4+\sqrt{11}\right)5}{16-11}+\dfrac{3-\sqrt{7}}{9-7}-\dfrac{6\left(\sqrt{7}+2\right)}{7-4}-\dfrac{\sqrt{7}-5}{2}\)

\(=4+\sqrt{11}-\dfrac{3-\sqrt{7}}{2}-2\left(\sqrt{7}+2\right)-\dfrac{\sqrt{7}-5}{2}=\dfrac{8+2\sqrt{11}-3+\sqrt{7}-4\sqrt{7}-8-\sqrt{7}+5}{2}=\dfrac{2\sqrt{11}-4\sqrt{7}+2}{2}=1+\sqrt{11}-2\sqrt{7}\)