Rút gọn biểu thức chứa căn bậc hai

$ \left( \sqrt{5}+\sqrt{2} \right)\sqrt{7-2\sqrt{10}} $

$ \begin{array}{l} =\left( \sqrt{5}+\sqrt{2} \right)\sqrt{5-2\sqrt{5}.\sqrt{2}+2}=\left( \sqrt{5}+\sqrt{2} \right)\sqrt{{{\left( \sqrt{5}-\sqrt{2} \right)}^{2}}} \\ =\left( \sqrt{5}+\sqrt{2} \right)\left| \sqrt{5}-\sqrt{2} \right|=\left( \sqrt{5}+\sqrt{2} \right)\left( \sqrt{5}-\sqrt{2} \right)=5-2=3. \end{array} $

Ta có

$ \begin{array}{l} A=\left( \dfrac{\sqrt{\text{x}}}{\sqrt{\text{x}}+3}+\dfrac{3}{\sqrt{\text{x}}-3} \right)\cdot \dfrac{\sqrt{\text{x}}+3}{\text{x}+9} \\ =\dfrac{\text{x}+9}{(\sqrt{\text{x}}+3)(\sqrt{\text{x}}-3)}\cdot \dfrac{\sqrt{\text{x}}+3}{\text{x}+9} \\ =\dfrac{1}{\sqrt{\text{x}}-3} \end{array} $

Ta có

$ \left( \dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}} \right):\dfrac{1}{a\left( \sqrt{7}-\sqrt{5} \right)}=\left( \dfrac{\sqrt{7}.\sqrt{2}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{5}.\sqrt{3}-\sqrt{5}}{1-\sqrt{3}} \right).a\left( \sqrt{7}-\sqrt{5} \right) $

$ =\left( \dfrac{-\sqrt{7}\left( 1-\sqrt{2} \right)}{1-\sqrt{2}}-\dfrac{\sqrt{5}\left( \sqrt{3}-1 \right)}{\sqrt{3}-1} \right).a\left( \sqrt{7}-\sqrt{5} \right)=\left[ \dfrac{\sqrt{6}\left( \sqrt{2}-1 \right)}{2\left( \sqrt{2}-1 \right)}-2\sqrt{6} \right].\left( -\dfrac{a}{\sqrt{6}} \right) $

$ =\left( -\sqrt{7}-\sqrt{5} \right).a\left( \sqrt{7}-\sqrt{5} \right)=-a.\left( \sqrt{7}+\sqrt{5} \right)\left( \sqrt{7}-\sqrt{5} \right)=-2a. $

Ta có $ x=3+2\sqrt{2}={{\left( \sqrt{2}+1 \right)}^{2}}\Rightarrow \sqrt{x}=\sqrt{{{\left( \sqrt{2}+1 \right)}^{2}}}=\sqrt{2}+1 $

Thay $ \sqrt{x}=\sqrt{2}+1 $ vào biểu thức $ P $ ta được

$ P=\dfrac{\sqrt{2}+1+1}{\sqrt{2}+1-2}=\dfrac{\sqrt{2}+2}{\sqrt{2}-1}=\dfrac{\left( \sqrt{2}+2 \right)\left( \sqrt{2}+1 \right)}{\left( \sqrt{2}-1 \right)\left( \sqrt{2}+1 \right)}=4+3\sqrt{2} $ .

Ta có $ A=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x} $ $ =\dfrac{\left( \sqrt{x}+1 \right)\left( \sqrt{x}+2 \right)+2\sqrt{x}\left( \sqrt{x}-2 \right)}{\left( \sqrt{x}-2 \right)\left( \sqrt{x}+2 \right)}-\dfrac{2+5\sqrt{x}}{\left( \sqrt{x}-2 \right)\left( \sqrt{x}+2 \right)} $

$ =\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left( \sqrt{x}-2 \right)\left( \sqrt{x}+2 \right)} $ $ =\dfrac{3x-6\sqrt{x}}{\left( \sqrt{x}-2 \right)\left( \sqrt{x}+2 \right)} $

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

Vậy $ A=\dfrac{3\sqrt{x}}{\sqrt{x}+2} $ với $ x\ge 0;x\ne 4 $ .

Ta có

$ \left( \sqrt{5}-1 \right)\sqrt{6+2\sqrt{5}}=\left( \sqrt{5}-1 \right)\sqrt{5+2\sqrt{5}.1+1}=\left( \sqrt{5}-1 \right)\sqrt{{{\left( \sqrt{5}+1 \right)}^{2}}} $

$ =\left( \sqrt{5}-1 \right)\left( \sqrt{5}+1 \right)=5-1=4 $ .

Ta có $ \begin{array}{l} P=2\Leftrightarrow \dfrac{x}{\sqrt{x}+1}=2\Leftrightarrow x-2\sqrt{x}-2=0\Leftrightarrow {{\left( \sqrt{x} \right)}^{2}}-2\sqrt{x}+1-3=0\Leftrightarrow {{\left( \sqrt{x}-1 \right)}^{2}}-3=0 \\ \Leftrightarrow \left[ \begin{array}{l} \sqrt{x}=\sqrt{3}+1\,\,\left( TM \right) \\ \sqrt{x}=-\sqrt{3}+1\,\left( L \right) \end{array} \right.\Rightarrow x={{\left( \sqrt{3}+1 \right)}^{2}}=4+2\sqrt{3}=\dfrac{2}{2-\sqrt{3}}. \end{array} $

Với $ x\ge 0 $ ta có $ P=\sqrt{x} $

$ \begin{array}{l} \Leftrightarrow \dfrac{3\sqrt{x}-1}{\sqrt{x}+1}=\sqrt{x}\Leftrightarrow \dfrac{3\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\sqrt{x}\left( \sqrt{x}+1 \right)}{\sqrt{x}+1} \\ \Rightarrow 3\sqrt{x}-1=x+\sqrt{x}\Leftrightarrow x-2\sqrt{x}+1=0 \\ \Leftrightarrow {{\left( \sqrt{x}-1 \right)}^{2}}=0\Leftrightarrow \sqrt{x}=1\Leftrightarrow x=1\left( TM \right). \end{array} $

Thay $ x=4 $ (thỏa điều kiện) vào $ P $ ta được $ P=\dfrac{\sqrt{4}}{\sqrt{4}-1}=\dfrac{2}{2-1}=2 $ .

$ 5\sqrt{a}+2\sqrt{\dfrac{a}{4}}-a\sqrt{\dfrac{4}{a}}-\sqrt{25a} $ $ =5\sqrt{a}+2.\dfrac{\sqrt{a}}{\sqrt{4}}-a\dfrac{\sqrt{4a}}{a}-5\sqrt{a} $

$ =5\sqrt{a}+\sqrt{a}-2\sqrt{a}-5\sqrt{a} $ $ =-\sqrt{a} $ .

Với điều kiện: $ x > 0,x\ne 4,x\ne 9 $ . Ta có: $ P=-1 $

$ \Leftrightarrow \dfrac{4x}{\sqrt{x}-3}=-1\Leftrightarrow 4x+\sqrt{x}-3=0\Leftrightarrow 4x+4\sqrt{x}-3\sqrt{x}-3=0 $

$ \Leftrightarrow 4\sqrt{x}(\sqrt{x}+1)-3(\sqrt{x}+1)=0 $

$ \Leftrightarrow (\sqrt{x}+1)(4\sqrt{x}-3)=0\Leftrightarrow \left[ \begin{array}{l} \sqrt{x}=-1(ktm) \\ \sqrt{x}=\dfrac{3}{4}\Leftrightarrow x=\dfrac{9}{16}(tm) \end{array} \right. $

Với $ x=\dfrac{9}{16} $ thì $ P=-1. $

Ta có

$ B=\left( \dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1} \right).\dfrac{{{(1-x)}^{2}}}{2} $ $ =\left( \dfrac{\sqrt{x}-2}{\left( \sqrt{x}-1 \right)\left( \sqrt{x}+1 \right)}-\dfrac{\sqrt{x}+2}{{{\left( \sqrt{x}+1 \right)}^{2}}} \right).\dfrac{{{(x-1)}^{2}}}{2} $ $ =\left( \dfrac{\left( \sqrt{x}-2 \right)\left( \sqrt{x}+1 \right)}{\left( \sqrt{x}-1 \right){{\left( \sqrt{x}+1 \right)}^{2}}}-\dfrac{\left( \sqrt{x}+2 \right)\left( \sqrt{x}-1 \right)}{\left( \sqrt{x}-1 \right){{\left( \sqrt{x}+1 \right)}^{2}}} \right).\dfrac{{{\left( \sqrt{x}-1 \right)}^{2}}{{\left( \sqrt{x}+1 \right)}^{2}}}{2} $ $ =\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left( \sqrt{x}-1 \right){{\left( \sqrt{x}+1 \right)}^{2}}}.\dfrac{{{\left( \sqrt{x}-1 \right)}^{2}}.{{\left( \sqrt{x}+1 \right)}^{2}}}{2} $

$ =\dfrac{-2\sqrt{x}\left( \sqrt{x}-1 \right)}{2}=\sqrt{x}-x $

Vậy $ B=\sqrt{x}-x $ .

Với $ x\ge 0;x\ne 4 $ ta có $ A=\dfrac{3\sqrt{x}}{\sqrt{x}+2} $

Xét $ A=2\Leftrightarrow \dfrac{3\sqrt{x}}{\sqrt{x}+2}=2\Rightarrow 3\sqrt{x}=2\left( \sqrt{x}+2 \right)\Leftrightarrow \sqrt{x}=4\Leftrightarrow x=16\left( TM \right) $

Vậy $ x=16 $ .

$ \sqrt{{{\left( 4-\sqrt{5} \right)}^{2}}}-\sqrt{6-2\sqrt{5}}=\sqrt{{{\left( 4-\sqrt{5} \right)}^{2}}}-\sqrt{5-2\sqrt{5}+1}=\sqrt{{{\left( 4-\sqrt{5} \right)}^{2}}}-\sqrt{{{\left( \sqrt{5}-1 \right)}^{2}}} $

$ =\left| 4-\sqrt{5} \right|-\left| \sqrt{5}-1 \right|=4-\sqrt{5}-\sqrt{5}+1=5-2\sqrt{5} $ .

$ \sqrt{{{\left( \sqrt{2}+\sqrt{5} \right)}^{2}}}-\sqrt{7-2\sqrt{10}}=\sqrt{{{\left( \sqrt{2}+\sqrt{5} \right)}^{2}}}-\sqrt{5-2\sqrt{5}.\sqrt{2}+2} $

$ =\sqrt{{{\left( \sqrt{2}+\sqrt{5} \right)}^{2}}}-\sqrt{{{\left( \sqrt{5}-\sqrt{2} \right)}^{2}}}=\left| \sqrt{2}+\sqrt{5} \right|-\left| \sqrt{5}-\sqrt{2} \right|=\sqrt{2}+\sqrt{5}-\sqrt{5}+\sqrt{2}=2\sqrt{2} $ .

Với $ a > 0 $ ta có $ 2\sqrt{a}-\sqrt{9{{a}^{3}}}+{{a}^{2}}\sqrt{\dfrac{16}{a}}+\dfrac{2}{{{a}^{2}}}\sqrt{36{{a}^{5}}} $

$ =2\sqrt{a}-\sqrt{9{{a}^{2}}.a}+{{a}^{2}}\dfrac{\sqrt{16a}}{a}+\dfrac{2}{{{a}^{2}}}.\sqrt{36{{a}^{4}}.a} $

$ =2\sqrt{a}-3a\sqrt{a}+4a\sqrt{a}+\dfrac{2}{{{a}^{2}}}.6{{a}^{2}}\sqrt{a} $ $ =2\sqrt{a}-3a\sqrt{a}+4a\sqrt{a}+12\sqrt{a}=14\sqrt{a}+a\sqrt{a} $ .

Ta có $ B=\dfrac{\sqrt{x}+3}{\sqrt{x}+2}=\dfrac{\left( \sqrt{x}+2 \right)+1}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}+2}=1+\dfrac{1}{\sqrt{x}+2} $

Vì $ x\ge 0\Leftrightarrow \sqrt{x}\ge 0\Rightarrow \sqrt{x}+2\ge 2 > 0 $ suy ra $ \dfrac{1}{\sqrt{x}+2} > 0\Leftrightarrow 1+\dfrac{1}{\sqrt{x+2}} > 1 $ hay $ B > 1 $ .

$ 3\sqrt{8a}+\dfrac{1}{4}\sqrt{\dfrac{32a}{25}}-\dfrac{a}{\sqrt{3}}.\sqrt{\dfrac{3}{2a}}-\sqrt{2a} $ $ =3\sqrt{4.2a}+\dfrac{1}{4}\dfrac{\sqrt{16.2a}}{\sqrt{25}}-\dfrac{a}{\sqrt{3}}.\dfrac{\sqrt{3}}{\sqrt{2a}}-\sqrt{2a} $

$ =3.2\sqrt{2a}+\dfrac{1}{4}.\dfrac{4\sqrt{2a}}{5}-\dfrac{a}{\sqrt{3}}.\dfrac{\sqrt{3}.\sqrt{2a}}{2a}-\sqrt{2a} $ $ =6\sqrt{2a}+\dfrac{1}{5}\sqrt{2a}-\dfrac{1}{2}\sqrt{2a}-\sqrt{2a} $

$ =\sqrt{2a}.\left( 6+\dfrac{1}{5}-\dfrac{1}{2}-1 \right)=\dfrac{47}{10}\sqrt{2a} $ .

Thu gọn biểu thức ta có $ B=\sqrt{x}-x $ Xét $ B > 0\Leftrightarrow \sqrt{x}-x > 0\Leftrightarrow \sqrt{x}\left( 1-\sqrt{x} \right) > 0 $

Với $ x\ge 0,x\ne 1 $ ta có $ \sqrt{x}\ge 0 $ nên $ \sqrt{x}(1-\sqrt{x}) > 0\Rightarrow \left\{ \begin{array}{l} 1-\sqrt{x} > 0 \\ x\ne 0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} \sqrt{x} < 1 \\ x\ne 0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} x < 1 \\ x\ne 0 \end{array} \right. $

Kết hợp điều kiện ta có $ 0 < x < 1 $ .

$ \begin{array}{l} A=\dfrac{{{\left( \sqrt{a} \right)}^{2}}-2\sqrt{ab}+{{\left( \sqrt{b} \right)}^{2}}+4\sqrt{ab}}{\sqrt{ab}}-\dfrac{\sqrt{{{a}^{2}}b}-\sqrt{a{{b}^{2}}}}{\sqrt{ab}} \\ =\dfrac{{{\left( \sqrt{a}+\sqrt{b} \right)}^{2}}}{\sqrt{a}+\sqrt{b}}-\dfrac{\sqrt{ab}\left( \sqrt{a}-\sqrt{b} \right)}{\sqrt{ab}} \\ =\sqrt{a}+\sqrt{b}-\left( \sqrt{a}-\sqrt{b} \right)=2\sqrt{b} \end{array} $

Ta có $ {{x}^{2}}-\sqrt{x}=\sqrt{x}\left( \sqrt{{{x}^{3}}}-1 \right)=\sqrt{x}\left( \sqrt{x}-1 \right)\left( x-\sqrt{x}+1 \right) $

Khi đó

$ \begin{array}{*{35}{l}} PT\Leftrightarrow \dfrac{\sqrt{x}+1}{\sqrt{x}\left( x+\sqrt{x}+1 \right)}\cdot \sqrt{x}\left( \sqrt{x}-1 \right)\left( x+\sqrt{x}+1 \right)=2 \\ \begin{array}{l} \Leftrightarrow \left( \sqrt{x}+1 \right)\left( \sqrt{x}-1 \right)=2 \\ \Leftrightarrow x-1=2 \\ \Leftrightarrow x=3 \end{array} \end{array} $

Ta có

$ \left( \dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3} \right).\left( \dfrac{-a}{\sqrt{6}} \right)=\left( \dfrac{2\sqrt{3}-\sqrt{2}.\sqrt{3}}{\sqrt{4.2}-2}-\dfrac{\sqrt{36.6}}{3} \right).\left( \dfrac{-a}{\sqrt{6}} \right) $ $ \begin{array}{l} =\left[ \dfrac{\sqrt{3}\left( 2-\sqrt{2} \right)}{2\sqrt{2}-2}-\dfrac{6\sqrt{6}}{3} \right].\left( -\dfrac{a}{\sqrt{6}} \right)=\left[ \dfrac{\sqrt{6}\left( \sqrt{2}-1 \right)}{2\left( \sqrt{2}-1 \right)}-2\sqrt{6} \right].\left( -\dfrac{a}{\sqrt{6}} \right) \\ =\left( \dfrac{\sqrt{6}}{2}-2\sqrt{6} \right).\left( \dfrac{-a}{\sqrt{6}} \right)=\left( -\dfrac{3\sqrt{6}}{2} \right).\left( \dfrac{-a}{\sqrt{6}} \right)=\dfrac{3a}{2}. \end{array} $

Ta có $ \left( \dfrac{1}{2}\sqrt{\dfrac{a}{2}}-\dfrac{3}{2}\sqrt{2a}+\dfrac{4}{5}\sqrt{200a} \right):\dfrac{1}{8}=\left( \dfrac{1}{2}.\dfrac{\sqrt{a}}{\sqrt{2}}-\dfrac{3}{2}\sqrt{2}.\sqrt{a}+\dfrac{4}{5}\sqrt{100}.\sqrt{2}.\sqrt{a} \right).8 $

$ =4.\dfrac{\sqrt{a}}{\sqrt{2}}-12\sqrt{2}.\sqrt{a}+\dfrac{32}{5}.10.\sqrt{2}.\sqrt{a}=4.\dfrac{\sqrt{2}.\sqrt{a}}{2}-12\sqrt{2}.\sqrt{a}+64\sqrt{2}.\sqrt{a} $

$ =2\sqrt{2a}-12\sqrt{2a}+64\sqrt{2a}=54\sqrt{2a} $ .

$ \sqrt{32}+\sqrt{50}-3\sqrt{8}-\sqrt{18}=\sqrt{16.2}+\sqrt{25.2}-3\sqrt{4.2}-\sqrt{9.2}=4\sqrt{2}+5\sqrt{2}-6\sqrt{2}-3\sqrt{2}=0 $ .

Ta có $ P=\dfrac{2.9}{\sqrt{9}+1}=\dfrac{18}{3+1}=\dfrac{18}{4}=\dfrac{9}{2}. $

Ta có

$ M=\dfrac{\sqrt{{{\left( x+2 \right)}^{2}}-8x}}{\sqrt{x}-\dfrac{2}{\sqrt{x}}}=\dfrac{\sqrt{{{\left( x-2 \right)}^{2}}}\cdot \sqrt{x}}{x-2}=\dfrac{\left| x-2 \right|\sqrt{x}}{x-2} $

Với $ 0 < x < 2\Rightarrow M=\dfrac{\left( 2-x \right)\sqrt{x}}{x-2}=-\dfrac{\left( x-2 \right)\sqrt{x}}{x-2}=-\sqrt{x} $

Ta có $ \dfrac{5}{12}-\dfrac{1}{\sqrt{6}}=\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}=\dfrac{5-2\sqrt{6}}{12}=\dfrac{{{\left( \sqrt{3}-\sqrt{2} \right)}^{2}}}{12} $

$ \begin{array}{l} \Rightarrow \dfrac{1}{\sqrt{3}}\sqrt{\dfrac{5}{12}-\dfrac{1}{\sqrt{6}}}=\dfrac{1}{\sqrt{3}}\dfrac{\sqrt{{{(\sqrt{3}-\sqrt{2})}^{2}}}}{12}=\dfrac{\sqrt{3}\cdot \sqrt{2}}{6} \\ \Rightarrow \text{A}=\left( \dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{6} \right)\cdot \dfrac{1}{\sqrt{3}} \\ =\left( \dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{3}-\sqrt{2}}{6} \right)\cdot \dfrac{1}{\sqrt{3}} \\ =\dfrac{3\sqrt{3}}{6}\cdot \dfrac{1}{\sqrt{3}}=\dfrac{1}{2} \end{array} $

Ta có $ C=\left( \dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{2}{x-\sqrt{x}} \right):\dfrac{1}{\sqrt{x}-1} $

$ $ $ \begin{array}{*{35}{l}} =\left( \dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{2}{\sqrt{x}\left( \sqrt{x}-1 \right)} \right).\left( \sqrt{x}-1 \right) \\ =\dfrac{x+2}{\sqrt{x}\left( \sqrt{x}-1 \right)}.\left( \sqrt{x}-1 \right) \\ =\dfrac{x+2}{\sqrt{x}} \end{array} $

Vậy $ C=\dfrac{x+2}{\sqrt{x}} $ với $ x > 0;x\ne 1 $

Điều kiện xác định: \[x \geqslant 0,x \ne 4\]

\begin{array} { l } { Q = \dfrac { ( \sqrt { x } + 1 ) ( \sqrt { x } + 2 ) - 2 \sqrt { x } ( \sqrt { x } - 2 ) - ( 5 \sqrt { x } + 2 ) } { ( \sqrt { x } - 2 ) ( \sqrt { x } + 2 ) } : \dfrac { \sqrt { x } ( 3 - \sqrt { x } ) } { ( \sqrt { x } + 2 ) ^ { 2 } } } \\ { = \dfrac { x + 3 \sqrt { x } + 2 - 2 x + 4 \sqrt { x } - 5 \sqrt { x } - 2 } { ( \sqrt { x } - 2 ) ( \sqrt { x } + 2 ) } \cdot \dfrac { ( \sqrt { x } + 2 ) ^ { 2 } } { \sqrt { x } ( 3 - \sqrt { x } ) } } \\ { = \dfrac { - x + 2 \sqrt { x } } { ( \sqrt { x } - 2 ) ( \sqrt { x } + 2 ) } \cdot \dfrac { ( \sqrt { x } + 2 ) ^ { 2 } } { \sqrt { x } ( 3 - \sqrt { x } ) } } \\ { = \dfrac { \sqrt { x } ( \sqrt { x } - 2 ) } { ( \sqrt { x } - 2 ) ( \sqrt { x } + 2 ) } \cdot \dfrac { ( \sqrt { x } + 2 ) ^ { 2 } } { \sqrt { x } ( 3 - \sqrt { x } ) } } \\ { = \dfrac { \sqrt { x } + 2 } { \sqrt { x } - 3 } } \end{array}

Khi đó 

\[\begin{gathered}
  Q = 2 \Rightarrow \dfrac{{\sqrt x  + 2}}{{\sqrt x  - 3}} = 2 \Rightarrow \sqrt x  + 2 = 2\left( {\sqrt x  - 3} \right) \hfill \\
   \Rightarrow \sqrt x  = 8 \Rightarrow x = 64 \hfill \\ 
\end{gathered} \]

Ta xét

$ P-4=\dfrac{x+2\sqrt{x}+2}{\sqrt{x}}-4=\dfrac{x+2\sqrt{x}+2-4\sqrt{x}}{\sqrt{x}}=\dfrac{x-2\sqrt{x}+2}{\sqrt{x}} $

$ =\dfrac{\left( x-2\sqrt{x}+1 \right)+1}{\sqrt{x}}=\dfrac{{{\left( \sqrt{x}-1 \right)}^{2}}+1}{\sqrt{x}} $

Vì $ {{\left( \sqrt{x}-1 \right)}^{2}}+1\ge 1 > 0,\forall x > 0 $ và $ \sqrt{x} > 0,\forall x > 0 $ nên $ P-4 > 0\Leftrightarrow P > 4 $ với $ x > 0 $ .

Ta có

$ \dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}+\dfrac{a-b}{\sqrt{a}+\sqrt{b}}=\dfrac{\sqrt{a}.\sqrt{a}.\sqrt{b}+\sqrt{b}.\sqrt{b}.\sqrt{a}}{\sqrt{ab}}+\dfrac{{{\left( \sqrt{a} \right)}^{2}}-{{\left( \sqrt{b} \right)}^{2}}}{\sqrt{a}+\sqrt{b}} $

$ =\dfrac{\sqrt{ab}\left( \sqrt{a}+\sqrt{b} \right)}{\sqrt{ab}}+\dfrac{\left( \sqrt{a}-\sqrt{b} \right)\left( \sqrt{a}+\sqrt{b} \right)}{\sqrt{a}+\sqrt{b}}=\sqrt{a}+\sqrt{b}+\sqrt{a}-\sqrt{b}=2\sqrt{a}. $

Ta có biểu thức A có nghĩa

$ \Leftrightarrow \left\{ \begin{array}{*{35}{l}} x\ge 0 \\ x\sqrt{x}+x+\sqrt{x}\ne 0 \\ {{x}^{2}}-\sqrt{x}\ne 0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{*{35}{l}} x\ge 0 \\ \sqrt{x}\left( x+\sqrt{x}+1 \right)\ne 0 \\ \sqrt{x}\left( \sqrt{x}-1 \right)\left( x+\sqrt{x}+1 \right)\ne 0 \end{array} \right.\Leftrightarrow 0 < x\ne 1 $

Ta có

$ \begin{array}{l} \dfrac{x-\sqrt{yz}}{\left( \sqrt{x}+\sqrt{y} \right)\left( \sqrt{x}+\sqrt{z} \right)}=\dfrac{x+\sqrt{xz}-\sqrt{xz}-\sqrt{yz}}{\left( \sqrt{x}+\sqrt{y} \right)\left( \sqrt{x}+\sqrt{z} \right)} \\ =\dfrac{\sqrt{x}\left( \sqrt{x}+\sqrt{z} \right)-\sqrt{z}\left( \sqrt{x}+\sqrt{z} \right)}{\left( \sqrt{x}+\sqrt{y} \right)\left( \sqrt{x}+\sqrt{z} \right)}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{z}} \end{array} $

$ \begin{array}{*{35}{l}} +)\dfrac{y-\sqrt{xz}}{\left( \sqrt{y}+\sqrt{x} \right)\left( \sqrt{y}+\sqrt{z} \right)}=\dfrac{\sqrt{y}}{\sqrt{y}+\sqrt{z}}-\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}} \\ +)\dfrac{z-\sqrt{xy}}{\left( \sqrt{z}+\sqrt{x} \right)\left( \sqrt{z}+\sqrt{y} \right)}=\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{z}}-\dfrac{\sqrt{y}}{\sqrt{y}+\sqrt{z}} \end{array} $

$ \Rightarrow A=0 $

Ta có

$ \begin{array}{l} \dfrac{a-b}{{{b}^{2}}}\sqrt{\dfrac{{{a}^{2}}{{b}^{4}}}{{{a}^{2}}-2ab+{{b}^{2}}}}=\dfrac{a-b}{{{b}^{2}}}.\dfrac{\sqrt{{{a}^{2}}{{b}^{4}}}}{\sqrt{{{(a-b)}^{2}}}}=\dfrac{(a-b)}{{{b}^{2}}}.\dfrac{\left| a \right|{{b}^{2}}}{\left| a-b \right|} \\ =\dfrac{(a-b)}{{{b}^{2}}}.\dfrac{\left| a \right|{{b}^{2}}}{(a-b)}=\left| a \right|. \end{array} $

Ta có $ P=\dfrac{2\sqrt{6}+\sqrt{3}+4\sqrt{2}+3}{\sqrt{11+2\left( \sqrt{6}+\sqrt{12}+\sqrt{18} \right)}}=\dfrac{\left( \sqrt{6}+3+3\sqrt{2} \right)+\left( \sqrt{2}+\sqrt{3}+\sqrt{6} \right)}{\sqrt{2+3+6+2\left( \sqrt{2}.\sqrt{3}+\sqrt{2}.\sqrt{6}+\sqrt{3}.\sqrt{6} \right)}} $

$ =\dfrac{\sqrt{3}\left( \sqrt{2}+\sqrt{3}+\sqrt{6} \right)+\left( \sqrt{2}+\sqrt{3}+\sqrt{6} \right)}{\sqrt{2+3+6+2\left( \sqrt{2}.\sqrt{3}+\sqrt{2}.\sqrt{6}+\sqrt{3}.\sqrt{6} \right)}}=\dfrac{\left( \sqrt{2}+\sqrt{3}+\sqrt{6} \right)\left( \sqrt{3}+1 \right)}{\sqrt{{{\left( \sqrt{2}+\sqrt{3}+\sqrt{6} \right)}^{2}}}} $

$ =\dfrac{\left( \sqrt{2}+\sqrt{3}+\sqrt{6} \right)\left( \sqrt{3}+1 \right)}{\sqrt{2}+\sqrt{3}+\sqrt{6}}=\sqrt{3}+1 $ .

Vậy $ P=\sqrt{3}+1 $ .

Ta có $ Q=\dfrac{x}{\sqrt{{{x}^{2}}-{{y}^{2}}}}-\left( 1+\dfrac{x}{\sqrt{{{x}^{2}}-{{y}^{2}}}} \right):\dfrac{y}{x-\sqrt{{{x}^{2}}-{{y}^{2}}}}=\dfrac{x}{\sqrt{{{x}^{2}}-{{y}^{2}}}}-\dfrac{x+\sqrt{{{x}^{2}}-{{y}^{2}}}}{\sqrt{{{x}^{2}}-{{y}^{2}}}}\cdot \dfrac{x-\sqrt{{{x}^{2}}-{{y}^{2}}}}{y} $

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

Vậy $ Q=\dfrac{\sqrt{x-y}}{\sqrt{x+y}} $ với $ x > y > 0 $

$ \begin{array}{l} P=x\sqrt{y}+y\sqrt{x}={{\left( \sqrt{x} \right)}^{2}}\sqrt{y}+{{\left( \sqrt{y} \right)}^{2}}\sqrt{x}=\sqrt{xy}\left( \sqrt{x}+\sqrt{y} \right) \\ Q=x\sqrt{x}+y\sqrt{y}={{\left( \sqrt{x} \right)}^{3}}+{{\left( \sqrt{y} \right)}^{3}}=\left( \sqrt{x}+\sqrt{y} \right)\left( x-\sqrt{xy}+y \right) \\ R=x-y={{\left( \sqrt{x} \right)}^{2}}-{{\left( \sqrt{y} \right)}^{2}}=\left( \sqrt{x}-\sqrt{y} \right)\left( \sqrt{x}+\sqrt{y} \right) \end{array} $

Vậy $ R=\left( \sqrt{x}-\sqrt{y} \right)\left( \sqrt{x}+\sqrt{y} \right) $

ĐKXĐ: $ \left\{ \begin{array}{l} x\ge 0 \\ x\ne 1 \\ x\ne 9 \end{array} \right. $

Ta có: $ P=\dfrac{\sqrt{x}}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+3}{\sqrt{x}-3}=1+\dfrac{3}{\sqrt{x}-3}. $

Để $ P $ nhận giá trị là số nguyên dương thì 

$ \left\{ \begin{array}{l} P\in Z \\ P > 0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} \dfrac{3}{\sqrt{x}-3}\in Z \\ 1+\dfrac{3}{\sqrt{x}-3} > 0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} \dfrac{3}{\sqrt{x}-3}\in Z \\ \dfrac{3}{\sqrt{x}-3} > -1 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} \dfrac{3}{\sqrt{x}-3}\in Z \\ \dfrac{3+\sqrt{x}-3}{\sqrt{x}-3} > 0 \end{array} \right. $

$ \begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l} \left( \sqrt{x}-3 \right)\in U(3)(1) \\ \dfrac{\sqrt{x}}{\sqrt{x}-3} > 0(2) \end{array} \right. \\ (1)\Leftrightarrow \left( \sqrt{x}-3 \right)\in \{1;3\} \end{array} $

$ \Leftrightarrow \left[ \begin{array}{l} \sqrt{x}-3=1 \\ \sqrt{x}-3=3 \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} \sqrt{x}=4 \\ \sqrt{x}=6 \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=16(tm) \\ x=36(tm) \end{array} \right. $

Nhận thấy với $ x=16;x=36 $ vẫn thỏa mãn (2).

Nên $ x=16 $ hoặc $ x=36 $ thì P nguyên dương.

Ta có: $ A=\dfrac{1}{\sqrt{3}-1}-\sqrt{27}+\dfrac{3}{\sqrt{3}}=\dfrac{\sqrt{3}+1}{\left( \sqrt{3}-1 \right)\left( \sqrt{3}+1 \right)}-\sqrt{9.3}+\dfrac{\sqrt{3}.\sqrt{3}}{\sqrt{3}} $

$ =\dfrac{\sqrt{3}+1}{2}-3\sqrt{3}+\sqrt{3}=\dfrac{\sqrt{3}+1-4\sqrt{3}}{2}=\dfrac{1-3\sqrt{3}}{2} $

và $ B=\dfrac{5+\sqrt{5}}{\sqrt{5}+2}+\dfrac{\sqrt{5}}{\sqrt{5}-1}-\dfrac{3\sqrt{5}}{3+\sqrt{5}} $

$ =\dfrac{\left( 5+\sqrt{5} \right)\left( \sqrt{5}-2 \right)}{\left( \sqrt{5}+2 \right)\left( \sqrt{5}-2 \right)}+\dfrac{\sqrt{5}\left( \sqrt{5}+1 \right)}{\left( \sqrt{5}-1 \right)\left( \sqrt{5}+1 \right)}-\dfrac{3\sqrt{5}\left( 3-\sqrt{5} \right)}{\left( 3+\sqrt{5} \right)\left( 3-\sqrt{5} \right)} $

$ =\dfrac{3\sqrt{5}-5}{1}+\dfrac{5+\sqrt{5}}{4}-\dfrac{9\sqrt{5}-15}{4}=\dfrac{12\sqrt{5}-20+5+\sqrt{5}-9\sqrt{5}+15}{4}=\sqrt{5} $

Ta thấy $ A=\dfrac{1-3\sqrt{3}}{2} < 0(do\,\,\,1-3\sqrt{3} < 0) $ và $ B=\sqrt{5} > 0 $ nên $ A < 0 < B $ .