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 $A < 0 < B$ .