Bài tập ôn tập chương 1

Vì $x = 25$ (TMĐK) nên ta có: $\sqrt x  = 5$

Khi đó ta có: $A = \dfrac{7}{{5 + 8}} = \dfrac{7}{{13}}$

$B = \dfrac{{\sqrt x }}{{\sqrt x  - 3}} + \dfrac{{2\sqrt x  - 24}}{{x - 9}}$   với $x \ge 0,{\rm{ }}x \ne 9$

$ = \dfrac{{\sqrt x (\sqrt x  + 3)}}{{(\sqrt x  - 3)(\sqrt x  + 3)}} + \dfrac{{2\sqrt x  - 24}}{{(\sqrt x  - 3)(\sqrt x  + 3)}}$

$\begin{array}{l} = \dfrac{{x + 3\sqrt x  + 2\sqrt x  - 24}}{{(\sqrt x  - 3)(\sqrt x  + 3)}}\\ = \dfrac{{x + 5\sqrt x  - 24}}{{(\sqrt x  - 3)(\sqrt x  + 3)}}\\ = \dfrac{{x - 3\sqrt x  + 8\sqrt x  - 24}}{{(\sqrt x  - 3)(\sqrt x  + 3)}}\\ = \dfrac{{\sqrt x (\sqrt x  - 3) + 8(\sqrt x  - 3)}}{{(\sqrt x  - 3)(\sqrt x  + 3)}}\\ = \dfrac{{(\sqrt x  - 3)(\sqrt x  + 8)}}{{(\sqrt x  - 3)(\sqrt x  + 3)}}\\ = \dfrac{{\sqrt x  + 8}}{{\sqrt x  + 3}}\end{array}$

Vậy \(B = \dfrac{{\sqrt x  + 8}}{{\sqrt x  + 3}}\,\,\) với $x \ge 0,{\rm{ }}x \ne 9$

$P = A.B$ nên ta có:$P = \dfrac{7}{{\sqrt x  + 8}}.\dfrac{{\sqrt x  + 8}}{{\sqrt x  + 3}} = \dfrac{7}{{\sqrt x  + 3}}$ với $ x \ge 0$;$x\ne 9$

+) Ta có $x \ge 0$ nên $P > 0$

+) $x \ge 0$  nên \(\sqrt x  + 3 \ge 3 \Leftrightarrow \dfrac{7}{{\sqrt x  + 3}} \le \dfrac{7}{3}\)

Nên : \(0 < P \le \dfrac{7}{3}\). Để \(P \in Z \Rightarrow P \in \left\{ {1;2} \right\}\)

+) $P = 1$ \(\dfrac{7}{{\sqrt x + 3}} = 1 \Rightarrow \sqrt x + 3 = 7 \Leftrightarrow \sqrt x = 4\)\( \Leftrightarrow x = 16\)  (thỏa mãn điều kiện)

+) $P = 2$ \(\dfrac{7}{{\sqrt x + 3}} = 2 \Rightarrow 2\sqrt x + 6 = 7 \Leftrightarrow \sqrt x = \dfrac{1}{2}\) \( \Leftrightarrow x = \dfrac{1}{4}\) (thỏa mãn điều kiện)

Vậy \(x \in \left\{ {\dfrac{1}{4};16} \right\}\)

Vậy có hai giá trị của \(x\) thỏa mãn đề bài.

Điều kiện \(x \ge 0;x \ne 9.\)

P =  $\left[ {\dfrac{{\sqrt x  + 1}}{{(\sqrt x  + 3)(\sqrt x  - 3)}} - \dfrac{{\sqrt x  - 3}}{{(\sqrt x  + 3)(\sqrt x  - 3)}}} \right]$$\left( {\sqrt x  - 3} \right)$

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

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

= $\dfrac{4}{{\sqrt x  + 3}}$

Vậy \(P = \dfrac{4}{{\sqrt x  + 3}}\)  với  \(x \ge 0;x \ne 9.\)

$\begin{array}{l}A = \left( {\dfrac{{\sqrt x }}{2} - \dfrac{1}{{2\sqrt x }}} \right)\left( {\dfrac{{x - \sqrt x }}{{\sqrt x  + 1}} - \dfrac{{x + \sqrt x }}{{\sqrt x  - 1}}} \right) = \dfrac{{x - 1}}{{2\sqrt x }}.\left( {\dfrac{{\sqrt x \left( {\sqrt x  - 1} \right)}}{{\sqrt x  + 1}} - \dfrac{{\sqrt x \left( {\sqrt x  + 1} \right)}}{{\sqrt x  - 1}}} \right)\\\,\,\,\,\,\, = \dfrac{{x - 1}}{{2\sqrt x }}.\dfrac{{\sqrt x {{\left( {\sqrt x  - 1} \right)}^2} - \sqrt x {{\left( {\sqrt x  + 1} \right)}^2}}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}} = \dfrac{{x - 1}}{{2\sqrt x }}.\dfrac{{\sqrt x \left[ {x - 2\sqrt x  + 1 - \left( {x + 2\sqrt x  + 1} \right)} \right]}}{{x - 1}}\\\,\,\,\,\,\, = \dfrac{{x - 2\sqrt x  + 1 - x - 2\sqrt x  - 1}}{2} = \dfrac{{ - 4\sqrt x }}{2} =  - 2\sqrt x .\end{array}$

Vậy \(A =  - 2\sqrt x \) với \(x > 0;\,\,x \ne 1.\)

$\begin{array}{l}B = \dfrac{x}{{x - 4}} + \dfrac{1}{{\sqrt x  - 2}} + \dfrac{1}{{\sqrt x  + 2}} = \dfrac{x}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}} + \dfrac{1}{{\sqrt x  - 2}} + \dfrac{1}{{\sqrt x  + 2}}\\\,\,\,\,\, = \dfrac{{x + \sqrt x  + 2 + \sqrt x  - 2}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}} = \dfrac{{x + 2\sqrt x }}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\\\,\,\,\,\, = \dfrac{{\sqrt x \left( {\sqrt x  + 2} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}} = \dfrac{{\sqrt x }}{{\sqrt x  - 2}}.\end{array}$

Vậy \(B = \dfrac{{\sqrt x }}{{\sqrt x  - 2}}\)  với \(x \ge 0;\,\,x \ne 4\).

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

Vậy \(D = \dfrac{2}{{\sqrt y }}\)  với \(x \ge 0;\,\,y \ge 0;\,\,xy \ne 1\)

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

Vậy \(M = 2x - \sqrt x \)  với \(x > 0.\)

$\begin{array}{l}P = \dfrac{{{a^2} + \sqrt a }}{{a - \sqrt a  + 1}} - \dfrac{{2a + \sqrt a }}{{\sqrt a }} + 1\\= \dfrac{{a.a + \sqrt a }}{{a - \sqrt a + 1}} - \dfrac{{2{{\left( {\sqrt a } \right)}^2} + \sqrt a }}{{\sqrt a }} + 1\\= \dfrac{{\sqrt a \left( {a\sqrt a  + 1} \right)}}{{a - \sqrt a  + 1}} - \dfrac{{\sqrt a \left( {2\sqrt a  + 1} \right)}}{{\sqrt a }} + 1\\  = \dfrac{{\sqrt a \left( {{{\left( {\sqrt a } \right)}^3} + 1} \right)}}{{a - \sqrt a + 1}} - \left( {2\sqrt a + 1} \right) + 1\\= \dfrac{{\sqrt a \left( {\sqrt a  + 1} \right)\left( {a - \sqrt a  + 1} \right)}}{{a - \sqrt a  + 1}} - \left( {2\sqrt a  + 1} \right) + 1\\= \sqrt a \left( {\sqrt a  + 1} \right) - 2\sqrt a  - 1 + 1\\ = a + \sqrt a  - 2\sqrt a \\= a - \sqrt a .\end{array}$

Vậy \(P = a - \sqrt a \)  với \(a > 0.\)

Điều kiện \(x > 0,x \ne 4\)

$\begin{array}{l}A = \left( {\dfrac{1}{{\sqrt x }} - \dfrac{{\sqrt x  - 1}}{{x + 2\sqrt x }}} \right):\left( {\dfrac{1}{{\sqrt x  + 2}} - \dfrac{{\sqrt x  + 1}}{{x - 4}}} \right)\\ = \left( {\dfrac{{\sqrt x  + 2}}{{\sqrt x \left( {\sqrt x  + 2} \right)}} - \dfrac{{\sqrt x  - 1}}{{\sqrt x \left( {\sqrt x  + 2} \right)}}} \right):\left( {\dfrac{{\sqrt x  - 2}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}} - \dfrac{{\sqrt x  + 1}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}} \right)\\ = \dfrac{3}{{\sqrt x \left( {\sqrt x  + 2} \right)}}:\dfrac{{ - 3}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}\\ = \dfrac{3}{{\sqrt x \left( {\sqrt x  + 2} \right)}}.\dfrac{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}{{ - 3}}\\ = \dfrac{{2 - \sqrt x }}{{\sqrt x }}\end{array}$

Vậy với \(x > 0,x \ne 4\)  thì  \(A = \dfrac{{2 - \sqrt x }}{{\sqrt x }}\).

Ta có:

\(\begin{array}{l}x = 9 - 4\sqrt 5  = {\left( {\sqrt 5 } \right)^2} - 2.2.\sqrt 5  + {2^2} = {\left( {\sqrt 5  - 2} \right)^2}\\ \Rightarrow \sqrt x  = \sqrt {{{\left( {\sqrt 5  - 2} \right)}^2}}  = \left| {\sqrt 5  - 2} \right| = \sqrt 5  - 2\left( {do\,\,\sqrt 5  - 2 > 0\,} \right)\end{array}\)

Khi đó ta có \(A = \dfrac{{2 - \sqrt x }}{{\sqrt x }} = \dfrac{{2 - \left( {\sqrt 5  - 2} \right)}}{{\sqrt 5  - 2}} = \dfrac{{4 - \sqrt 5 }}{{\sqrt 5  - 2}} = \dfrac{{\left( {4 - \sqrt 5 } \right)\left( {\sqrt 5  + 2} \right)}}{{{{\left( {\sqrt 5 } \right)}^2} - {2^2}}} = \dfrac{{4\sqrt 5  + 8 - 5 - 2\sqrt 5 }}{1} = 2\sqrt 5  + 3\)

Vậy với \(x = 9 - 4\sqrt 5 \) thì \(A = 2\sqrt 5  + 3\)

$A{\rm{ }} < \;0$ \( \Leftrightarrow \) \(A = \dfrac{{2 - \sqrt x }}{{\sqrt x }} < 0\)

Với \(x > 0,x \ne 4\) ta có: \(\sqrt x  > 0\) .

Để \(\dfrac{{2 - \sqrt x }}{{\sqrt x }} < 0\) thì  $2 - \sqrt x  < 0 \Leftrightarrow \sqrt x  > 2 \Leftrightarrow x > 4$

Vậy ta có: \(x > 4\)  thì \(A < 0.\)

Ta có: với \(x > 0,x \ne 1\)

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

Vậy với \(x > 0,x \ne 1\) ta có  \(P = \dfrac{{\sqrt x  + 1}}{{\sqrt x }}\)

Để \(2P = 2\sqrt x  + 5\)

Ta có: với \(x > 0,x \ne 1\)

\(\begin{array}{l}2P = 2\sqrt x  + 5\\ \Leftrightarrow 2.\dfrac{{\sqrt x  + 1}}{{\sqrt x }} = 2\sqrt x  + 5\\ \Rightarrow 2\sqrt x  + 2 = 2x + 5\sqrt x \\ \Leftrightarrow 2x + 3\sqrt x  - 2 = 0\\ \Leftrightarrow 2x - \sqrt x  + 4\sqrt x  - 2 = 0\\ \Leftrightarrow \sqrt x \left( {2\sqrt x  - 1} \right) + 2\left( {2\sqrt x  - 1} \right) = 0\\ \Leftrightarrow \left( {2\sqrt x  - 1} \right)\left( {\sqrt x  + 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sqrt x  = \dfrac{1}{2} \Leftrightarrow x = \dfrac{1}{4}(tm)\\\sqrt x  =  - 2(ktm)\end{array} \right.\end{array}\)

Vậy \(x = \dfrac{1}{4}\) thì \(2P = 2\sqrt x  + 5\)

$A = \dfrac{{2x}}{{x + 3\sqrt x  + 2}} + \dfrac{{5\sqrt x  + 1}}{{x + 4\sqrt x  + 3}} + \dfrac{{\sqrt x  + 10}}{{x + 5\sqrt x  + 6}}$$= \dfrac{{2x}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)}} + \dfrac{{5\sqrt x  + 1}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 3} \right)}} + \dfrac{{\sqrt x  + 10}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  + 3} \right)}}$

\( = \dfrac{{2x\left( {\sqrt x  + 3} \right) + \left( {5\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right) + \left( {\sqrt x  + 10} \right)\left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)\left( {\sqrt x  + 3} \right)}}\)

\( = \dfrac{{2x\sqrt x  + 6x + 5x + 11\sqrt x  + 2 + x + 11\sqrt x  + 10}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)\left( {\sqrt x  + 3} \right)}}\)

\( = \dfrac{{2x\sqrt x  + 12x + 22\sqrt x  + 12}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)\left( {\sqrt x  + 3} \right)}}\)

\( = \dfrac{{2x\sqrt x  + 2x + 10x + 10\sqrt x  + 12\sqrt x  + 12}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)\left( {\sqrt x  + 3} \right)}}\)

\(\begin{array}{l} = \dfrac{{2x\left( {\sqrt x  + 1} \right) + 10\sqrt x \left( {\sqrt x  + 1} \right) + 12\left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)\left( {\sqrt x  + 3} \right)}}\\ = \dfrac{{\left( {\sqrt x  + 1} \right)\left( {2x + 10\sqrt x  + 12} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)\left( {\sqrt x  + 3} \right)}}\\
= \dfrac{{2\left( {x + 5\sqrt x + 6} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2\left( {x + 2\sqrt x + 3\sqrt x + 6} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2\left[ {\sqrt x \left( {\sqrt x + 2} \right) + 3\left( {\sqrt x + 2} \right)} \right]}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x + 3} \right)}}\\= \dfrac{{2\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)\left( {\sqrt x  + 3} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 2} \right)\left( {\sqrt x  + 3} \right)}}\\ = 2\end{array}\)

Vậy giá trị của $A$ không phụ thuộc vào biến $x.$

Điều kiện xác định: \(x \ne 1;x > 0\)

\(\begin{array}{l}P = 1:\left( {\dfrac{{x + 2}}{{x\sqrt x  - 1}} + \dfrac{{\sqrt x  + 1}}{{x + \sqrt x  + 1}} - \dfrac{{\sqrt x  + 1}}{{x - 1}}} \right)\\ = 1:\left( {\dfrac{{x + 2}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}} + \dfrac{{\sqrt x  + 1}}{{\left( {x + \sqrt x  + 1} \right)}} - \dfrac{{\sqrt x  + 1}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}}} \right)\\ = 1:\dfrac{{\left( {x + 2} \right)\left( {\sqrt x  + 1} \right) + {{\left( {\sqrt x  + 1} \right)}^2}\left( {\sqrt x  - 1} \right) - \left( {\sqrt x  + 1} \right)\left( {x + \sqrt x  + 1} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}\end{array}\)

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

\(\begin{array}{l} = \dfrac{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}{{x\sqrt x  - \sqrt x }}\\ = \dfrac{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}{{\sqrt x \left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}}\\ = \dfrac{{x + \sqrt x  + 1}}{{\sqrt x }}\end{array}\)

Vậy \(P = \dfrac{{x + \sqrt x  + 1}}{{\sqrt x }}\)  với \(x \ne 1;x > 0\)

+ So sánh \(P\) với \(3.\)

Xét \(P - 3 = \dfrac{{x + \sqrt x  + 1}}{{\sqrt x }} - 3 = \dfrac{{x + \sqrt x  + 1 - 3\sqrt x }}{{\sqrt x }} = \dfrac{{{{\left( {\sqrt x  - 1} \right)}^2}}}{{\sqrt x }}\)

Với \(x \ne 1;x > 0\) ta có: \(\sqrt x  > 0\); $\sqrt x \ne 1$ nên \({\left( {\sqrt x  - 1} \right)^2} > 0\) suy ra: $P-3 > 0$ hay $P > 3$

ĐKXĐ: \(\left\{ \begin{array}{l}x \ge 0\\\sqrt x  - 3 \ne 0\\x - 9 \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\x \ne 9\end{array} \right..\)

$\begin{array}{l}P = \left( {\dfrac{{2\sqrt x }}{{\sqrt x  + 3}} + \dfrac{{\sqrt x }}{{\sqrt x  - 3}} - \dfrac{{3x + 3}}{{x - 9}}} \right):\left( {\dfrac{{2\sqrt x  - 2}}{{\sqrt x  - 3}} - 1} \right)\\\,\,\,\,\, = \dfrac{{2\sqrt x \left( {\sqrt x  - 3} \right) + \sqrt x \left( {\sqrt x  + 3} \right) - 3x - 3}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}:\dfrac{{2\sqrt x  - 2 - \sqrt x  + 3}}{{\sqrt x  - 3}}\\\,\,\,\,\, = \dfrac{{2x - 6\sqrt x  + x + 3\sqrt x  - 3x - 3}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}:\dfrac{{\sqrt x  + 1}}{{\sqrt x  - 3}}\\\,\,\,\,\, = \dfrac{{ - 3\sqrt x  - 3}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}.\dfrac{{\sqrt x  - 3}}{{\sqrt x  + 1}}\\\,\,\,\,\, = \dfrac{{ - 3\left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  + 1} \right)}} = \dfrac{{ - 3}}{{\sqrt x  + 3}}.\end{array}$

Vậy \(P = \dfrac{{ - 3}}{{\sqrt x  + 3}}\)  với \(x \ge 0;x \ne 9.\)

Ta có: $x = \dfrac{{3 - \sqrt 5 }}{2} = \dfrac{{6 - 2\sqrt 5 }}{4} = \dfrac{{{{\left( {\sqrt 5  - 1} \right)}^2}}}{4}$

\(\begin{array}{l} \Rightarrow \sqrt x  = \sqrt {\dfrac{{{{\left( {\sqrt 5  - 1} \right)}^2}}}{4}}  = \dfrac{{\left| {\sqrt 5  - 1} \right|}}{2} = \dfrac{{\sqrt 5  - 1}}{2}.\\ \Rightarrow P = \dfrac{{ - 3}}{{\dfrac{{\sqrt 5  - 1}}{2} + 3}} = \dfrac{{ - 3.2}}{{\sqrt 5  - 1 + 6}} = \dfrac{{ - 6}}{{\sqrt 5  + 5}} = \dfrac{{ - 6\left( {5 - \sqrt 5 } \right)}}{{{5^2} - 5}} = \dfrac{{6\sqrt 5  - 30}}{{20}} = \dfrac{{3\sqrt 5  - 15}}{{10}}.\end{array}\)

Ta có \(P = \dfrac{{ - 3}}{{\sqrt x  + 3}}\)  với \(x \ge 0;x \ne 9.\) Suy ra

$\begin{array}{l}P <  - \dfrac{1}{2} \Leftrightarrow  - \dfrac{3}{{\sqrt x  + 3}} <  - \dfrac{1}{2}\\ \Leftrightarrow \dfrac{3}{{\sqrt x  + 3}} > \dfrac{1}{2} \Leftrightarrow \dfrac{3}{{\sqrt x  + 3}} - \dfrac{1}{2} > 0\\ \Leftrightarrow \dfrac{{6 - \sqrt x  - 3}}{{2\left( {\sqrt x  + 3} \right)}} > 0\\ \Leftrightarrow 3 - \sqrt x  > 0\,\,\,\left( {do\,\,\,\sqrt x  + 3 > 0\,\,\forall x \ge 0;\,x \ne 9} \right)\\ \Leftrightarrow \sqrt x  < 3 \Leftrightarrow x < 9.\end{array}$

Kết hợp với ĐKXĐ ta được với \(0 \le x < 9\) thì $P <  - \dfrac{1}{2}$.

Ta có \(P = \dfrac{{ - 3}}{{\sqrt x  + 3}}\)  với \(x \ge 0;x \ne 9.\)

+ Với \(x\)  không là số chính phương thì \(\sqrt x \)  là số vô tỉ nên \(P = \dfrac{{ - 3}}{{\sqrt x  + 3}}\) là số vô tỉ (loại)

+ Với \(x\) là số chính phương

Ta có:

$\begin{array}{l}P \in Z \Leftrightarrow \dfrac{{ - 3}}{{\sqrt x  + 3}} \in Z \Leftrightarrow \left( {\sqrt x  + 3} \right) \in U\left( { - 3} \right) \Leftrightarrow \left( {\sqrt x  + 3} \right) \in \left\{ {1;\,3} \right\}\,\,\left( {do\,\,\sqrt x  + 3 > 0\,\,\forall x \ge 0;\,x \ne 9} \right).\\ \Leftrightarrow \left[ \begin{array}{l}\sqrt x  + 3 = 1\\\sqrt x  + 3 = 3\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\sqrt x  =  - 2\,\left( {ktm} \right)\\\sqrt x  = 0\end{array} \right. \Leftrightarrow x = 0\left( {tm} \right)\end{array}$

Vậy x = 0 thì $P \in Z$.

ĐKXĐ: \(\left\{ \begin{array}{l}x \ge 0\\y \ge 0\\\sqrt {xy}  + 2\sqrt x  - 3\sqrt y  - 6 \ne 0\\\sqrt {xy}  + 2\sqrt x  + 3\sqrt y  + 6 \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\y \ge 0\\\left( {\sqrt y  + 2} \right)\left( {\sqrt x  - 3} \right) \ne 0\\\left( {\sqrt y  + 2} \right)\left( {\sqrt x  + 3} \right) \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\y \ge 0\\\sqrt x  \ne 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\y \ge 0\\x \ne 9\end{array} \right..\)

\(\begin{array}{l}K = \dfrac{{2\sqrt x  + 3\sqrt y }}{{\sqrt {xy}  + 2\sqrt x  - 3\sqrt y  - 6}} - \dfrac{{6 - \sqrt {xy} }}{{\sqrt {xy}  + 2\sqrt x  + 3\sqrt y  + 6}}\\ = \dfrac{{2\sqrt x  + 3\sqrt y }}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt y  + 2} \right)}} - \dfrac{{6 - \sqrt {xy} }}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt y  + 2} \right)}}\\ = \dfrac{{\left( {2\sqrt x  + 3\sqrt y } \right)\left( {\sqrt x  + 3} \right) - \left( {6 - \sqrt {xy} } \right)\left( {\sqrt x  - 3} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt y  + 2} \right)\left( {\sqrt x  + 3} \right)}}\\ = \dfrac{{2x + 6\sqrt x  + 3\sqrt {xy}  + 9\sqrt y  - \left( {6\sqrt x  - 18 - x\sqrt y  + 3\sqrt {xy} } \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt y  + 2} \right)\left( {\sqrt x  + 3} \right)}}\\ = \dfrac{{2x + x\sqrt y  + 9\sqrt y  + 18}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt y  + 2} \right)\left( {\sqrt x  + 3} \right)}} = \dfrac{{\left( {\sqrt y  + 2} \right)\left( {x + 9} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt y  + 2} \right)\left( {\sqrt x  + 3} \right)}}\\ = \dfrac{{x + 9}}{{x - 9}}.\end{array}\)

Vậy \(K = \dfrac{{x + 9}}{{x - 9}}\)  với \(x;y \ge 0;x \ne 9.\)

Ta có \(K = \dfrac{{x + 9}}{{x - 9}}\)  với \(x;y \ge 0;x \ne 9.\) Nên

 \(\begin{array}{l}K = \dfrac{{y + 81}}{{y - 81}} \Rightarrow \dfrac{{x + 9}}{{x - 9}} = \dfrac{{y + 81}}{{y - 81}}\\ \Rightarrow  \left( {x + 9} \right)\left( {y - 81} \right) = \left( {y + 81} \right)\left( {x - 9} \right)\\ \Leftrightarrow xy + 9y - 81x - 9.81 = xy - 9y + 81x - 9.81\\ \Leftrightarrow 9y = 81x\\ \Leftrightarrow \dfrac{y}{x} = \dfrac{{81}}{9} = 9.\end{array}\)

Vậy nếu \(K = \dfrac{{y + 81}}{{y - 81}}\) thì \(\dfrac{y}{x}\) là số nguyên chia hết cho 3.